Chapterwise Revision

If for the value of any transaction, discount = x %.

It's discounted value = (100 - x)% of its original value.

MRP (in ₹)	Discount %	Discounted value (Selling price)
600	40	60% of 600 = ₹ 360
450	32	68% of 450 = ₹ 306
900	20	80% of 900 = ₹ 720
750	30	70% of 750 = ₹ 525
Total		₹ 1911

CGST = 6% of ₹ 1911 = $\frac{6}{100} \times 1938$ = ₹ 114.66

SGST = CGST = ₹ 114.66

Amount of bill = ₹ 1911 + ₹ 114.66 + ₹ 114.66 = ₹ 2140.32

Hence, amount of bill = ₹ 2140.32

If for the value of any transaction, discount = x %.

It's discounted value = (100 - x)% of its original value.

MRP (in ₹)	Discount %	Discounted value (Selling price)
9600	20	80% of 9600 = ₹ 7680
6000	50	50% of 6000 = ₹ 3000
10800	40	60% of 10800 = ₹ 6480
9000	30	70% of 9000 = ₹ 6300
7200	40	60% of 7200 = ₹ 4320
Total		₹ 27780

IGST = 5% of ₹ 27780 = $\frac{5}{100} \times 27780 = \frac{138900}{100}$ = ₹ 1389

Amount of bill = ₹ 27780 + ₹ 1389 = ₹ 29169

Hence, amount of bill = ₹ 29169.

This is a interstate transaction.

∴ GST = IGST.

MRP (in ₹)	Discount %	Discounted value (Selling price)	GST %	IGST	Amount (in ₹)
12000	30	70% of 12000 = ₹8400	18	18% of ₹8400 = ₹1512	8400 + 1512 = 9912
15000	40	60% of 15000 = ₹9000	18	18% of ₹9000 = ₹1620	9000 + 1620 = 10620
5200	30	70% of 5200 = ₹3640	12	12% of ₹3640 = ₹436.80	3640 + 436.80 = 4076.80
8000	40	60% of 8000 = ₹4800	12	12% of ₹4800 = ₹576	4800 + 576 = 5376
Total					29984.80

Hence, the amount of bill = ₹ 29984.80

This is an intra-state transaction.

GST = CGST + SGST

We know, SGST = CGST

MRP (in ₹)/ item	Number of items	Total cost of whole quantity (in ₹)	GST %	CGST (in ₹)	SGST (in ₹)	Amount (MRP + CGST + SGST) (in ₹)
300	40	12000	12	6% of 12000 = 720	720	12000 + 720 + 720 = 13440
600	50	30000	12	6% of 30000 = 1800	1800	30000 + 1800 + 1800 = 33600
480	70	33600	18	9% of 33600 = 3024	3024	33600 + 3024 + 3024 = 39648
Total						₹ 86688

Hence, amount of bill = ₹ 86688.

When product is sold from Kanpur to Meerut (It is an intra-state transaction)

For dealer in Kanpur :

S.P. = ₹ 5000

CGST = 9% of ₹ 5000 = $\frac{9}{100} \times 5000 = ₹ 450$.

SGST = CGST = ₹ 450.

When product is sold from Meerut to Delhi (It is an inter-state transaction)

For dealer in Meerut :

Input-tax credit (ITC) = ₹ 450 + ₹ 450 = ₹ 900.

C.P. (excluding GST) = ₹ 5000

Profit = ₹ 2000

S.P. = ₹ 5000 + ₹ 2000 = ₹ 7000

IGST = 18% of ₹ 7000 = $\frac{18}{100} \times 7000$ = ₹ 1260

C.P. in Delhi = S.P. in Meerut + IGST = ₹ 7000 + ₹ 1260 = ₹ 8260.

Hence, cost of goods/services in Delhi = ₹ 8260.

Given,

P = ₹ 3200

r = 9%

Maturity value of recurring deposit = Total sum deposited + Interest on it

= P × n + P × $\frac{n(n + 1)}{2 \times 12} \times \frac{r}{100}$

where n is time in months.

Time (n) = 3 years = 3 × 12 = 36 months.

Substituting values we get :

$$\begin{aligned}= 3200 \times 36 + 3200 \times \frac{36 \times (36 + 1)}{2 \times 12} \times \frac{9}{100} \\[1em] = 115200 + 3200 \times \frac{36 \times 37}{24} \times \frac{9}{100} \\[1em] = 115200 + \frac{115200 \times 37 \times 9}{2400} \\[1em] = 115200 + 48 \times 37 \times 9 \\[1em] = 115200 + 15984 \\[1em] = 131184.\end{aligned}$$

Hence, maturity value = ₹ 131184.

(i) Given,

Time (n) = 3 years = 3 × 12 = 36 months.

Rate = 8%

Let monthly instalment be ₹ P.

By formula,

Interest = P × $\frac{n(n + 1)}{2 \times 12} \times \frac{r}{100}$

Substituting values we get

$$\begin{aligned}\Rightarrow 9990 = P \times \frac{36 \times 37}{2 \times 12} \times \frac{8}{100} \\[1em] \Rightarrow 9990 = P \times \frac{3 \times 37}{2} \times \frac{8}{100} \\[1em] \Rightarrow 9990 = P \times \frac{3 \times 37}{25} \\[1em] \Rightarrow P = \frac{9990 \times 25}{3 \times 37} \\[1em] \Rightarrow P = \frac{249750}{111} \\[1em] \Rightarrow P = ₹ 2250.\end{aligned}$$

Hence, monthly installment = ₹ 2250.

(ii) Maturity value of recurring deposit = Total sum deposited + Interest on it

= P × n + 9990

= 2250 × 36 + 9990

= 81000 + 9990

= ₹ 90990.

Hence, the maturity value of this account = ₹ 90990.

Let rate of interest be r%.

Given,

P = ₹ 240

M.V. = ₹ 17694

Time (n) = 5 years = 5 × 12 = 60 months.

Maturity value = P × n + P × $\frac{n(n + 1)}{2 \times 12} \times \frac{r}{100}$

Substituting values we get :

⇒ 17694 = 240 × 60 + 240 × $\frac{60 \times 61}{2 \times 12} \times \frac{r}{100}$

⇒ 17694 = 14400 + 6 × $\frac{305r}{5}$

⇒ 17694 - 14400 = 6 × 61r

⇒ 3294 = 366r

⇒ r = $\frac{3294}{366}$ = 9%.

Hence, rate of interest = 9%.

Let time = n months.

Given,

P = ₹ 2000

r = 10%

By formula,

M.V. = P × n + P × $\frac{n(n + 1)}{2 \times 12} \times \frac{r}{100}$

Substituting values we get :

$$\begin{aligned}\Rightarrow 83100 = 2000 \times n + 2000 \times \frac{n(n + 1)}{2 \times 12} \times \frac{10}{100} \\[1em] \Rightarrow 83100 = 2000n + \frac{100n(n + 1)}{12} \\[1em] \Rightarrow 83100 = \frac{24000n + 100n^2 + 100n}{12} \\[1em] \Rightarrow 83100 \times 12 = 100(240n + n^2 + n) \\[1em] \Rightarrow 241n + n^2 = \frac{83100 \times 12}{100} \\[1em] \Rightarrow 241n + n^2 = 9972 \\[1em] \Rightarrow n^2 + 241n - 9972 = 0 \\[1em] \Rightarrow n^2 + 277n - 36n - 9972 = 0\\[1em] \Rightarrow n(n + 277) - 36(n + 277) = 0 \\[1em] \Rightarrow (n - 36)(n + 277) = 0 \\[1em] \Rightarrow n - 36 = 0 \text{ or } n + 277 = 0 \\[1em] \Rightarrow n = 36 \text{ or } n = -277.\end{aligned}$$

Since, time cannot be negative.

∴ n = 36 months = 3 years.

Hence, time = 3 years.

Let rate of interest be r%.

Given,

P = ₹ 900

Time (n) = 2 years = 24 months.

By formula,

Interest = P × $\frac{n(n + 1)}{2 \times 12} \times \frac{r}{100}$

Substituting values we get :

$$\begin{aligned}\Rightarrow 1800 = 900 \times \frac{24 \times 25}{24} \times \frac{r}{100} \\[1em] \Rightarrow 1800 = 225r \\[1em] \Rightarrow r = \frac{1800}{225} \\[1em] \Rightarrow r = 8%.\end{aligned}$$

Hence, the rate of interest = 8%.

Given,

Nominal value (N.V.) = ₹ 100

Income (dividend) on one share = $4\frac{1}{2}$ % of N.V. = $\frac{4\frac{1}{2}}{100} \times 100 = 4\frac{1}{2}$.

Given,

Total income = ₹ 72

⇒ No. of shares × Income on one share = 72

⇒ No. of shares × 4.5 = 72

⇒ No. of shares = $\frac{72}{4.5} = 16$.

We know that,

⇒ Sum invested = M.V. of each share × No. of shares

⇒ 1800 = M.V. of each share × 16

⇒ M.V. of each share = $\frac{1800}{16}$ = ₹ 112.50

Hence, M.V. of each share = ₹ 112.50

(i) Given,

Nominal value (N.V.) = ₹ 100

Income (dividend) on one share = 10 % of N.V. = $\frac{10}{100} \times 100 = 10$.

Given,

Total income = ₹ 800

⇒ No. of shares × Income on one share = 800

⇒ No. of shares × 10 = 800

⇒ No. of shares = $\frac{800}{10} = 80$.

We know that,

⇒ Sum invested = M.V. of each share × No. of shares

⇒ 10000 = M.V. of each share × 80

⇒ M.V. of each share = $\frac{10000}{80}$ = ₹ 125

Hence, M.V. of each share = ₹ 125

(ii) Given,

Man earns an income of ₹ 800 on investing ₹ 10000.

Rate percent = $\frac{\text{Income}}{\text{Investment}} \times 100$

$$\begin{aligned}= \frac{800}{10000} \times 100 \\[1em] = \frac{80000}{10000} \\[1em] = 8%.\end{aligned}$$

Hence, person earns 8% on his investment.

(i) Given,

A man holds 800 shares.

Nominal Value (N.V.) of each share = ₹ 100

Dividend paid semi-annually = 7.5%

Dividend paid annually = 7.5% × 2 = 15%

(i) Dividend on each share = 15% of N.V.

= $\frac{15}{100} \times 100$ = ₹ 15.

Total dividend = Dividend on each share × No. of shares

= ₹ 15 × 800 = ₹ 12000.

Hence, annual dividend of person = ₹ 12000.

(ii) Given,

The person had bought the shares at 40% premium.

Market Value (MV) = N.V. + Premium = 100 + $\frac{40}{100} \times 100$ = 100 + 40 = ₹ 140.

Total investment = M.V. × No. of shares

= 140 × 800 = ₹ 112000.

Rate of return = $\frac{\text{Income}}{\text{Investment}} \times 100 = \frac{12000}{112000} \times 100 = \frac{1200}{112}$ = 10.714 %.

Hence, there is a 10.714% percentage return on investment.

Given,

N.V. of share = ₹ 100

M.V. = N.V. + Premium = ₹ 100 + ₹ 32 = ₹ 132.

(i) No. of shares bought = $\frac{\text{Investment}}{\text{M.V.}} = \frac{10560}{132}$ = 80.

Hence, no. of shares = 80.

(ii) Income on 1 share = 9% of N.V.

= $\frac{9}{100} \times 100$ = ₹ 9.

Annual income = Income on 1 share × No. of shares

= ₹ 9 × 80 = ₹ 720.

Hence, annual income = ₹ 720.

(iii) Rate of return = $\frac{\text{Income}}{\text{Investment}} \times 100$

$$\begin{aligned}= \frac{720}{10560} \times 100 \\[1em] = \frac{72000}{10560} \\[1em] = 6\frac{9}{11}%.\end{aligned}$$

Hence, rate of return = $6\frac{9}{11}$ %.

Given,

Nominal value (N.V.) = ₹ 25

Income (dividend) on one share = 12 % of N.V. = $\frac{12}{100} \times 25 = 3$.

Given,

Total income = ₹ 1875

⇒ No. of shares × Income on one share = 1875

⇒ No. of shares × 3 = 1875

⇒ No. of shares = $\frac{1875}{3} = 625$.

We know that,

⇒ Sum invested = M.V. of each share × No. of shares

⇒ 20000 = M.V. of each share × 625

⇒ M.V. of each share = $\frac{20000}{625}$ = ₹ 32

Hence, M.V. of each share = ₹ 32.

(i) M.V. = ₹ 26 + $\frac{10}{100} \times 26$ = ₹ 26 + ₹ 2.6 = ₹ 28.6

Investment = ₹ 20,020

No. of shares bought = $\frac{20020}{28.6}$ = 700.

Hence, no. of shares bought = 700.

(ii) Annual dividend = No. of shares × Rate of div. × N.V. of 1 share

= 700 × $\frac{15}{100} \times 26$

= ₹ 2,730.

Hence, dividend = ₹ 2,730.

(iii) Rate of interest = $\frac{2730}{20020} \times 100 = \frac{273000}{20020} = 13\frac{7}{11}$%.

Hence, the rate of interest = $13\frac{7}{11}$%.

(i) M.V. of first type of share = ₹ 100 + ₹ 20 = ₹ 120.

Investment = ₹ 9,600

No. of shares = $\frac{\text{Investment}}{\text{M.V.}} = \frac{9600}{120}$ = 80.

Hence, no. of shares = 80.

(ii) S.P. of 1 share = ₹ 160,

S.P. of 80 shares = 80 × ₹ 160 = ₹ 12,800.

Hence, sale proceeds = ₹ 12,800.

(iii) M.V. of second type of share = ₹ 40.

Investment = ₹ 12,800

No. of shares = $\frac{\text{Investment}}{\text{M.V.}} = \frac{12800}{40}$ = 320.

Hence, no. of shares = 320.

(iv) Annual dividend = No. of shares × Rate of div. × N.V. of 1 share

In first case :

Annual dividend = 80 × $\frac{8}{100} \times 100$ = ₹ 640.

In second case :

Annual dividend = 320 × $\frac{10}{100} \times 50$ = ₹ 1600.

Difference = ₹ 1600 - ₹ 640 = ₹ 960

Hence, there is an increase of ₹ 960 in dividend amount.

(i) From graph,

A = {x ∈ R : -3 < x ≤ 2}, B = {x ∈ R : 0 ≤ x < 5}

(ii) From graph,

A ∪ B = {x ∈ R : -3 < x < 5}

A ∩ B = {x ∈ R : 0 ≤ x ≤ 2}

A' ∩ B = {x ∈ R : -2 < x < 5}

A - B = {x ∈ R : -3 < x < 0}

B - A = {x ∈ R : 2 < x < 5}

Solving L.H.S. of the above inequation :

$$\begin{aligned}\Rightarrow -2 \le \frac{1}{2} - \frac{2x}{3} \\[1em] \Rightarrow \frac{2x}{3} \le \frac{1}{2} + 2 \\[1em] \Rightarrow \frac{2x}{3} \le \frac{5}{2} \\[1em] \Rightarrow x \le \frac{5}{2} \times \frac{3}{2} \\[1em] \Rightarrow x \le \frac{15}{4} \\[1em] \Rightarrow x \le 3\frac{3}{4}\text{ ...........(1)}\end{aligned}$$

Solving R.H.S. of the inequation :

$$\begin{aligned}\Rightarrow \frac{1}{2} - \frac{2x}{3} \lt 1\frac{5}{6} \\[1em] \Rightarrow \frac{1}{2} - \frac{2x}{3} \lt \frac{11}{6} \\[1em] \Rightarrow \frac{1}{2} - \frac{11}{6} \lt \frac{2x}{3} \\[1em] \Rightarrow \frac{2x}{3} \gt \frac{3 - 11}{6} \\[1em] \Rightarrow \frac{2x}{3} \gt -\frac{8}{6} \\[1em] \Rightarrow \frac{2x}{3} \gt -\frac{4}{3} \\[1em] \Rightarrow x \gt -\frac{4}{3} \times \frac{3}{2} \\[1em] \Rightarrow x \gt -2 \text{ .........(2)}\end{aligned}$$

From (1) and (2),

-2 < x ≤ $3\frac{3}{4}$

Since, x ∈ N

∴ x = {1, 2, 3}.

Hence, x = {1, 2, 3}.

(a) False
Reason — If x < a and x < b, then (x - a)(x - b) will be greater than 0 as multiplication of two negative number generates a positive number.

(b) True
Reason — If a < 0 and b < 0, then (a + b) will be a negative number and square of any negative number is greater than 0, so (a + b)² is greater than 0.

(c) False
Reason — If a > b, then a² > b² is true only if a and b are positive numbers.

(d) True
Reason — If a positive number is added to any number (suppose x), then resultant number is greater than x.

(e) False
Reason — If, a < b and both are two negative integers then on recripocating the numbers the sign will be reversed, so $\frac{1}{a} \gt \frac{1}{b}$.

Solving,

⇒ 20 - 5x < 5(x + 8)

⇒ 20 - 5x < 5x + 40

⇒ 5x + 5x > 20 - 40

⇒ 10x > -20

⇒ x > $-\frac{20}{10}$ = -2.

(i) As, x > -2 and x ∈ I.

∴ x = {-1, 0, 1 .......}

The smallest no. = -1.

Hence, smallest value of x = -1, when x is an integer.

(ii) As, x > -2 and x ∈ W.

∴ x = {0, 1, 2 .......}

The smallest no. = 0.

Hence, smallest value of x = 0, when x is a whole number.

(iii) As, x > -2 and x ∈ N.

∴ x = {1, 2 .......}

The smallest no. = 1.

Hence, smallest value of x = 1, when x is a natural number.

Given,

⇒ 2 + 4x < 2x - 5 ≤ 3x

Solving L.H.S. of the above inequation :

⇒ 2 + 4x < 2x - 5

⇒ 4x - 2x < - 5 - 2

⇒ 2x < -7

⇒ x < $-\frac{7}{2}$

⇒ x < -3.5 ..........(1)

Solving R.H.S. of the above inequation :

⇒ 2x - 5 ≤ 3x

⇒ 3x - 2x ≥ -5

⇒ x ≥ -5 .............(2)

From (1) and (2), we get :

-5 ≤ x < -3.5; Since, x ∈ Z

Hence, x = {-5, -4}.

Solving 7x - 4 > 5x + 2,

⇒ 7x - 5x > 2 + 4

⇒ 2x > 6

⇒ x > 3.

P = {x : x > 3, x ∈ R}

Solving x - 19 ≥ 1 - 3x

⇒ x + 3x ≥ 1 + 19

⇒ 4x ≥ 20

⇒ x ≥ 5.

Q = {x : x ≥ 5, x ∈ R}

P ∩ Q = Numbers common between P and Q

= {x : x ≥ 5, x ∈ R}

P ∩ Q on the number line is :

Given,

$-2\frac{5}{6} \lt \frac{1}{2} - \frac{2x}{3} \le 2$

Solving L.H.S. of the inequation,

$$\begin{aligned}\Rightarrow -2\frac{5}{6} \lt \frac{1}{2} - \frac{2x}{3} \\[1em] \Rightarrow -\frac{17}{6} \lt \frac{1}{2} - \frac{2x}{3} \\[1em] \Rightarrow \frac{2x}{3} \lt \frac{1}{2} + \frac{17}{6} \\[1em] \Rightarrow \frac{2x}{3} \lt \frac{3 + 17}{6} \\[1em] \Rightarrow \frac{2x}{3} \lt \frac{20}{6} \\[1em] \Rightarrow x \lt \frac{20}{6} \times \frac{3}{2} \\[1em] \Rightarrow x \lt 5 .......(i)\end{aligned}$$

Solving R.H.S. of the inequation,

$$\begin{aligned}\Rightarrow \frac{1}{2} - \frac{2x}{3} \le 2 \\[1em] \Rightarrow \frac{2x}{3} \ge \frac{1}{2} - 2 \\[1em] \Rightarrow \frac{2x}{3} \ge -\frac{3}{2} \\[1em] \Rightarrow x \ge -\frac{3}{2} \times \frac{3}{2} \\[1em] \Rightarrow x \ge -\frac{9}{4} \\[1em] \Rightarrow x \ge -2.25 ........(ii)\end{aligned}$$

From (i) and (ii) we get,

-2.25 ≤ x < 5

Since, x ∈ W,

∴ Solution set = {0, 1, 2, 3, 4}.

Solution on the number line is :

Solving :

$$\begin{aligned}\Rightarrow \frac{8}{x + 3} - 2 = \frac{3}{2 - x} \\[1em] \Rightarrow \frac{8 - 2(x + 3)}{x + 3} = \frac{3}{2 - x} \\[1em] \Rightarrow \frac{8 - 2x - 6}{x + 3} = \frac{3}{2 - x} \\[1em] \Rightarrow \frac{2 - 2x}{x + 3} = \frac{3}{2 - x} \\[1em] \Rightarrow (2 - 2x)(2 - x) = 3(x + 3) \\[1em] \Rightarrow 4 - 2x - 4x + 2x^2 = 3x + 9 \\[1em] \Rightarrow 2x^2 - 6x + 4 = 3x + 9 \\[1em] \Rightarrow 2x^2 - 6x - 3x + 4 - 9 = 0 \\[1em] \Rightarrow 2x^2 - 9x - 5 = 0 \\[1em] \Rightarrow 2x^2 - 10x + x - 5 = 0 \\[1em] \Rightarrow 2x(x - 5) + 1(x - 5) = 0 \\[1em] \Rightarrow (2x + 1)(x - 5) = 0 \\[1em] \Rightarrow 2x + 1 = 0 \text{ or } x - 5 = 0 \\[1em] \Rightarrow x = -\frac{1}{2} \text{ or } x = 5.\end{aligned}$$

Hence, x = $-\frac{1}{2}$ or 5.

Solving :

$$\begin{aligned}\Rightarrow \frac{x(6 - x) + 9}{3(6 - x)} = \frac{2(6 + x)}{15} \\[1em] \Rightarrow \frac{6x - x^2 + 9}{18 - 3x} = \frac{12 + 2x}{15} \\[1em] \Rightarrow 15(6x - x^2 + 9) = (12 + 2x)(18 - 3x) \\[1em] \Rightarrow 90x - 15x^2 + 135 = 216 - 36x + 36x - 6x^2 \\[1em] \Rightarrow 90x - 15x^2 + 135 = 216 - 6x^2 \\[1em] \Rightarrow 15x^2 - 6x^2 + 216 - 135 - 90x = 0 \\[1em] \Rightarrow 9x^2 + 81 - 90x = 0 \\[1em] \Rightarrow 9(x^2 + 9 - 10x) = 0 \\[1em] \Rightarrow x^2 - 10x + 9 = 0 \\[1em] \Rightarrow x^2 - x - 9x + 9 = 0 \\[1em] \Rightarrow x(x - 1) - 9(x - 1) = 0 \\[1em] \Rightarrow (x - 9)(x - 1) = 0 \\[1em] \Rightarrow x - 9 = 0 \text{ or } x - 1 = 0 \\[1em] \Rightarrow x = 9 \text{ or } x = 1.\end{aligned}$$

Hence, x = 1 or 9.

Given,

Equation : k²x² - 2(2k - 1)x + 4 = 0

The roots of the equation are real and equal, when D = 0.

⇒ b² - 4ac = 0

⇒ [-2(2k - 1)]² - 4 × k² × 4 = 0

⇒ [-4k + 2]² - 16k² = 0

⇒ 16k² + 4 - 16k - 16k² = 0

⇒ 16k = 4

⇒ k = $\frac{4}{16} = \frac{1}{4}$.

Hence, k = $\frac{1}{4}$.

Solving :

$$\begin{aligned}\Rightarrow \frac{x}{a} - \frac{a + b}{x} = \frac{b(a + b)}{ax} \\[1em] \Rightarrow \frac{x^2 - a(a + b)}{ax} = \frac{ab + b^2}{ax} \\[1em] \Rightarrow \frac{x^2 - a(a + b)}{ax} \times ax = ab + b^2 \\[1em] \Rightarrow x^2 - a^2 - ab = ab + b^2 \\[1em] \Rightarrow x^2 = a^2 + ab + ab + b^2 \\[1em] \Rightarrow x^2 = a^2 + 2ab + b^2 \\[1em] \Rightarrow x^2 = (a + b)^2 \\[1em] \Rightarrow x = \sqrt{(a + b)^2} \\[1em] \Rightarrow x = (a + b), -(a + b).\end{aligned}$$

Hence, x = (a + b), -(a + b).

As, -5 is the root of the quadratic equation 2x² + px - 15 = 0, so it will satisfy the equation.

∴ 2(-5)² + (-5)p - 15 = 0

⇒ 2 × 25 - 5p - 15 = 0

⇒ 50 - 5p - 15 = 0

⇒ 5p = 35

⇒ p = $\frac{35}{5}$ = 7.

Substituting value of p in p(x² + x) + k = 0, we get :

⇒ 7(x² + x) + k = 0

⇒ 7x² + 7x + k = 0

Since, above equation has real and equal roots.

∴ D = 0

⇒ b² - 4ac = 0

⇒ (7)² - 4 × 7 × k = 0

⇒ 49 - 28k = 0

⇒ 28k = 49

⇒ k = $\frac{49}{28} = \frac{7}{4} = 1\frac{3}{4}$.

Hence, k = $1\frac{3}{4}$.

Since,

x articles are bought at ₹ (x - 8) each and (x - 2) some other articles are bought at ₹ (x - 3) each.

Total cost = x(x - 8) + (x - 2)(x - 3).

Given,

Total cost = ₹ 76.

∴ x(x - 8) + (x - 2)(x - 3) = 76

⇒ x² - 8x + x² - 3x - 2x + 6 = 76

⇒ 2x² - 13x + 6 - 76 = 0

⇒ 2x² - 13x - 70 = 0

⇒ 2x² - 20x + 7x - 70 = 0

⇒ 2x(x - 10) + 7(x - 10) = 0

⇒ (2x + 7)(x - 10) = 0

⇒ 2x + 7 = 0 or x - 10 = 0

⇒ 2x = -7 or x = 10

⇒ x = -$\frac{7}{2}$ or x = 10.

Since, no. of articles cannot be negative.

∴ x = 10.

Hence, there are 10 articles of first kind.

Let ten's digit be x.

Unit's digit = x + 2

Number = 10x + x + 2 = 11x + 2

Sum of digits = x + x + 2 = 2x + 2

According to question,

Number × Sum of digits = 144

⇒ (11x + 2)(2x + 2) = 144

⇒ 22x² + 22x + 4x + 4 = 144

⇒ 22x² + 26x + 4 = 144

⇒ 22x² + 26x = 140

⇒ 2(11x² + 13x) = 140

⇒ 11x² + 13x = 70

⇒ 11x² + 13x - 70 = 0

⇒ 11x² + 35x - 22x - 70 = 0

⇒ x(11x + 35) - 2(11x + 35) = 0

⇒ (x - 2)(11x + 35) = 0

⇒ x - 2 = 0 or 11x + 35 = 0

⇒ x = 2 or x = -$\frac{35}{11}$

Since, digit cannot be negative.

∴ x = 2.

Number = 11x + 2 = 11(2) + 2

= 22 + 2 = 24.

Hence, number = 24.

Let speed while going be x km/hour and speed while returning will be (x + 10) km/hour.

Time taken while going = $\frac{\text{Distance}}{\text{Time}} = \frac{150}{x}$ hours.

Time taken while returning = $\frac{\text{Distance}}{\text{Time}} = \frac{150}{x + 10}$ hours.

Given,

Time taken by a person to cover 150 km was 2.5 hours more than the time taken in return journey.

$$\begin{aligned}\therefore \frac{150}{x} - \frac{150}{x + 10} = 2.5 \\[1em] \Rightarrow \frac{150(x + 10) - 150x}{x(x + 10)} = \frac{25}{10} \\[1em] \Rightarrow \frac{150x + 1500 - 150x}{x(x + 10)} = \frac{5}{2} \\[1em] \Rightarrow \frac{1500}{x^2 + 10x} = \frac{5}{2} \\[1em] \Rightarrow 3000 = 5(x^2 + 10x) \\[1em] \Rightarrow x^2 + 10x = \frac{3000}{5} \\[1em] \Rightarrow x^2 + 10x = 600 \\[1em] \Rightarrow x^2 + 10x - 600 = 0 \\[1em] \Rightarrow x^2 + 30x - 20x - 600 = 0 \\[1em] \Rightarrow x(x + 30) - 20(x + 30) = 0 \\[1em] \Rightarrow (x - 20)(x + 30) = 0 \\[1em] \Rightarrow x - 20 = 0 \text{ or } x + 30 = 0 \\[1em] \Rightarrow x = 20 \text{ or } x = -30.\end{aligned}$$

Since, speed cannot be negative.

∴ x = 20 km/hour and (x + 10) = 30 km/hour.

Hence, speed while going = 20 km/hour and speed while returning = 30 km/hour.

Let A alone takes x days and B alone takes y days.

Given,

A takes 9 days more than B to do a certain piece of work

∴ x = y + 9 .........(1)

As, A takes x days to do a work

So, one day work of A = $\frac{1}{x}$

As, B takes y days to do a work

So, one day work of B = $\frac{1}{y}$

As, together they can do work in 6 days.

∴ In one day both of them do $\frac{1}{6}$ th part of work.

$$\begin{aligned}\therefore \frac{1}{x} + \frac{1}{y} = \frac{1}{6} \\[1em] \Rightarrow \frac{1}{y + 9} + \frac{1}{y} = \frac{1}{6} \space [\text{From (1)}] \\[1em] \Rightarrow \frac{y + y + 9}{y(y + 9)} = \frac{1}{6} \\[1em] \Rightarrow \frac{2y + 9}{y^2 + 9y} = \frac{1}{6} \\[1em] \Rightarrow 6(2y + 9) = y^2 + 9y \\[1em] \Rightarrow 12y + 54 = y^2 + 9y \\[1em] \Rightarrow y^2 + 9y - 12y - 54 = 0 \\[1em] \Rightarrow y^2 - 3y - 54 = 0\\[1em] \Rightarrow y^2 - 9y + 6y - 54 = 0 \\[1em] \Rightarrow y(y - 9) + 6(y - 9) = 0 \\[1em] \Rightarrow (y + 6)(y - 9) = 0 \\[1em] \Rightarrow y + 6 = 0 \text{ or } y - 9 = 0 \\[1em] \Rightarrow y = -6 \text{ or } y = 9.\end{aligned}$$

Since, days cannot be negative so, y ≠ -6.

x = y + 9 = 18.

Hence, A alone will take 18 days to complete the work.

Given,

Man bought a certain number of chairs for ₹ 10000.

Let no. of chairs bought be x.

∴ Cost of one chair = ₹ $\frac{10000}{x}$

S.P. of each chair = ₹ $\frac{10000}{x} + 50$

Since, he kept one chair for himself so he sold (x - 1) chairs.

Given,

⇒ Profit = ₹ 450

⇒ S.P. - C.P. = ₹ 450

$$\begin{aligned}\Rightarrow (x - 1) \times \Big(\frac{10000}{x} + 50\Big) - 10000 = 450 \\[1em] \Rightarrow 10000 + 50x - \frac{10000}{x} - 50 - 10000 = 450 \\[1em] \Rightarrow 50x - \frac{10000}{x} = 450 + 50 \\[1em] \Rightarrow \frac{50x^2 - 10000}{x} = 500 \\[1em] \Rightarrow 50x^2 - 10000 = 500x \\[1em] \Rightarrow 50(x^2 - 200) = 500x \\[1em] \Rightarrow x^2 - 200 = 10x \\[1em] \Rightarrow x^2 - 10x - 200 = 0 \\[1em] \Rightarrow x^2 - 20x + 10x - 200 = 0 \\[1em] \Rightarrow x(x - 20) + 10(x - 20) = 0 \\[1em] \Rightarrow (x + 10)(x - 20) = 0 \\[1em] \Rightarrow x + 10 = 0 \text{ or } x - 20 = 0 \\[1em] \Rightarrow x = -10 \text{ or } x = 20.\end{aligned}$$

Since no. of chairs cannot be negative.

∴ x = 20.

Hence, no. of chairs bought = 20.

From figure,

Length of larger portion (L) = (80 + x) meters

Breadth of larger portion (B) = (50 + x) meters

Area of larger portion = L × B = (80 + x)(50 + x)

= 4000 + 80x + 50x + x²

= (x² + 130x + 4000) m².

Length of smaller portion (l) = 80 meters

Breadth of smaller portion (b) = 50 meters

Area of smaller portion = l × b = 80 × 50 = 4000 m²

Area of unshaded portion = Area of larger portion - Area of smaller portion

= x² + 130x + 4000 - 4000

= (x² + 130x) m².

Given,

Area of unshaded portion is 75% of the area of the shaded portion.

$$\begin{aligned}\therefore 75% \text{Area of Shaded portion = Area of unshaded portion} \\[1em] \Rightarrow \frac{75}{100} \times 4000 = x^2 + 130x \\[1em] \Rightarrow 3000 = x^2 + 130x \\[1em] \Rightarrow x^2 + 130x - 3000 = 0 \\[1em] \Rightarrow x^2 + 150x - 20x - 3000 = 0 \\[1em] \Rightarrow x(x + 150) - 20(x + 150) = 0 \\[1em] \Rightarrow (x - 20)(x + 150) = 0 \\[1em] \Rightarrow x - 20 = 0 \text{ or } x + 150 = 0 \\[1em] \Rightarrow x = 20 \text{ or } x = -150.\end{aligned}$$

Since, length cannot be negative.

∴ x = 20 meters.

Hence, x = 20 meters.

Applying componendo and dividendo, we get :

$$\begin{aligned}\Rightarrow \frac{\sqrt{a + x} + \sqrt{a - x} + \sqrt{a + x} - \sqrt{a - x}}{(\sqrt{a + x} + \sqrt{a - x}) - (\sqrt{a + x} - \sqrt{a - x})} = \frac{b + 1}{b - 1} \\[1em] \Rightarrow \frac{2\sqrt{a + x}}{2\sqrt{a - x}} = \frac{b + 1}{b - 1} \\[1em] \Rightarrow \frac{\sqrt{a + x}}{\sqrt{a - x}} = \frac{b + 1}{b - 1} \\[1em]\end{aligned}$$

Squaring both sides we get :

$\Rightarrow \frac{a + x}{a - x} = \frac{(b + 1)^2}{(b - 1)^2}$

Again applying componendo and dividendo, we get :

$$\begin{aligned}\Rightarrow \frac{a + x + a - x}{a + x - (a - x)} = \frac{(b + 1)^2 + (b - 1)^2}{(b + 1)^2 - (b - 1)^2} \\[1em] \Rightarrow \frac{a + a + x - x}{a - a + x - (-x)} = \frac{b^2 + 1 + 2b + b^2 + 1 - 2b}{b^2 + 1 + 2b - (b^2 + 1 - 2b)} \\[1em] \Rightarrow \frac{2a}{2x} = \frac{2(b^2 + 1)}{4b} \\[1em] \Rightarrow \frac{a}{x} = \frac{b^2 + 1}{2b} \\[1em] \Rightarrow x = \frac{2ab}{b^2 + 1}.\end{aligned}$$

Hence, x = $\frac{2ab}{b^2 + 1}$.

Given,

a : b = 2 : 3

$$\begin{aligned}\Rightarrow \frac{a}{b} = \frac{2}{3} \\[1em] \Rightarrow \frac{a}{b} = \frac{2 \times 8}{3 \times 8} \\[1em] \Rightarrow \frac{a}{b} = \frac{16}{24} ...........(1)\end{aligned}$$

b : c = 4 : 5

$$\begin{aligned}\Rightarrow \frac{b}{c} = \frac{4}{5} \\[1em] \Rightarrow \frac{b}{c} = \frac{4 \times 6}{5 \times 6} \\[1em] \Rightarrow \frac{b}{c} = \frac{24}{30} ...........(2)\end{aligned}$$

c : d = 6 : 7

$$\begin{aligned}\Rightarrow \frac{c}{d} = \frac{6}{7} \\[1em] \Rightarrow \frac{c}{d} = \frac{6 \times 5}{7 \times 5} \\[1em] \Rightarrow \frac{c}{d} = \frac{30}{35} ...........(3)\end{aligned}$$

From (1), (2) and (3), we get :

a : b : c : d = 16 : 24 : 30 : 35

Hence, a : b : c : d = 16 : 24 : 30 : 35.

Given,

$\frac{\sqrt{2a + 1} + \sqrt{2a - 1}}{\sqrt{2a + 1} - \sqrt{2a - 1}}$ = x.

Applying componendo and dividendo, we get :

$$\begin{aligned}\Rightarrow \frac{\sqrt{2a + 1} + \sqrt{2a - 1} + \sqrt{2a + 1} - \sqrt{2a - 1}}{\sqrt{2a + 1} + \sqrt{2a - 1} - (\sqrt{2a+ 1} - \sqrt{2a - 1})} = \frac{x + 1}{x - 1} \\[1em] \Rightarrow \frac{2\sqrt{2a + 1}}{2\sqrt{2a - 1}} = \frac{x + 1}{x - 1} \\[1em] \Rightarrow \frac{\sqrt{2a + 1}}{\sqrt{2a - 1}} = \frac{x + 1}{x - 1}\end{aligned}$$

Squaring both sides we get,

$\Rightarrow \frac{2a + 1}{2a - 1} = \frac{(x + 1)^2}{(x - 1)^2}$

Applying componendo and dividendo we get :

$$\begin{aligned}\Rightarrow \frac{2a + 1 + 2a - 1}{2a + 1 - (2a - 1)} = \frac{(x + 1)^2 + (x - 1)^2}{(x + 1)^2 - (x - 1)^2} \\[1em] \Rightarrow \frac{4a}{2} = \frac{x^2 + 1 + 2x + x^2 + 1 - 2x}{x^2 + 1 + 2x - (x^2 + 1 - 2x)} \\[1em] \Rightarrow 2a = \frac{2(x^2 + 1)}{4x} \\[1em] \Rightarrow 8ax = 2(x^2 + 1) \\[1em] \Rightarrow 4ax = x^2 + 1 \\[1em] \Rightarrow x^2 - 4ax + 1 = 0.\end{aligned}$$

Hence, proved that x² - 4ax + 1 = 0.

(i) Calculating the compounded ratio :

$$\begin{aligned}\Rightarrow \frac{a - b}{a + b} \times \frac{b^2 + ab}{a^2 - ab} \\[1em] \Rightarrow \frac{a - b}{a + b} \times \frac{b(a + b)}{a(a - b)} \\[1em] \Rightarrow \frac{b}{a} \\[1em] \Rightarrow b : a.\end{aligned}$$

Hence, compounded ratio of (a - b) : (a + b) and (b² + ab) : (a² - ab) = b : a.

(ii) Calculating the compounded ratio :

$$\begin{aligned}\Rightarrow \frac{x + y}{x - y} \times \frac{x^2 + y^2}{(x + y)^2} \times \frac{(x^2 - y^2)^2}{(x^4 - y^4)} \\[1em] \Rightarrow \frac{1}{x - y} \times \frac{x^2 + y^2}{(x + y)} \times \frac{(x^2 - y^2)^2}{(x^2 - y^2)(x^2 + y^2)} \\[1em] \Rightarrow \frac{1}{x - y} \times \frac{\cancel{x^2 + y^2}}{(x + y)} \times \frac{(x^2 - y^2)^{\cancel{2}}}{\cancel{(x^2 - y^2)}\cancel{(x^2 + y^2)}} \\[1em] \Rightarrow \frac{(x^2 - y^2)}{(x - y)(x + y)} \\[1em] \Rightarrow \frac{(x^2 - y^2)}{(x^2 - y^2)} \\[1em] \Rightarrow 1 : 1.\end{aligned}$$

Hence, the compounded ratio = 1 : 1.

(iii) Calculating the compounded ratio :

$$\begin{aligned}\Rightarrow \frac{x^2 - 25}{x^2 + 3x - 10} \times \frac{x^2 - 4}{x^2 + 3x + 2} \times \frac{x + 1}{x^2 + 2x} \\[1em] \Rightarrow \frac{x^2 - 25}{x^2 + 5x - 2x - 10} \times \frac{(x - 2)(x + 2)}{x^2 + x + 2x + 2} \times \frac{x + 1}{x(x + 2)} \\[1em] \Rightarrow \frac{(x - 5)(x + 5)}{x(x + 5) - 2(x + 5)} \times \frac{(x - 2)(x + 2)}{x(x + 1) + 2(x + 1)} \times \frac{x + 1}{x(x + 2)} \\[1em] \Rightarrow \frac{(x - 5)(x + 5)}{(x - 2)(x + 5)} \times \frac{(x - 2)(x + 2)}{(x + 1)(x + 2)} \times \frac{x + 1}{x(x + 2)} \\[1em] \Rightarrow \frac{(x - 5)\cancel{(x + 5)}}{\cancel{(x - 2)}\cancel{(x + 5)}} \times \frac{\cancel{(x - 2)}\cancel{(x + 2)}}{\cancel{(x + 1)}\cancel{(x + 2)}} \times \frac{\cancel{x + 1}}{x(x + 2)} \\[1em] \Rightarrow \frac{(x - 5)}{x(x + 2)} \\[1em] \Rightarrow (x - 5) : x(x + 2).\end{aligned}$$

Hence, the compounded ratio = (x - 5) : x(x + 2).

Let price of first fan be f₁ and second fan be f₂.

Given,

Ratio of the prices of first two fans was 16 : 23.

$$\begin{aligned}\Rightarrow \frac{f_1}{f_2} = \frac{16}{23} \\[1em] \Rightarrow f_1 = \frac{16f_2}{23} ..........(1)\end{aligned}$$

First fan price after 2 years = $f_1 + \frac{10}{100}f_1 = f_1 + \frac{f_1}{10} = \frac{11f_1}{10}$.

Second fan price after two years = $f_2 + 477$.

Given,

Ratio of fan's prices becomes 11 : 20 after 2 years.

$$\begin{aligned}\therefore \frac{\frac{11}{10}f_1}{f_2 + 477} = \frac{11}{20} \\[1em] \Rightarrow \frac{11f_1}{10(f_2 + 477)} = \frac{11}{20} \\[1em] \Rightarrow \frac{f_1}{f_2 + 477} = \frac{11}{20} \times \frac{10}{11} \\[1em] \Rightarrow \frac{f_1}{f_2 + 477} = \frac{1}{2} \\[1em] \Rightarrow f_1 = \frac{f_2 + 477}{2}\end{aligned}$$

Substituting value of f₁ from (1) in above equation we get :

$$\begin{aligned}\Rightarrow \frac{16f_2}{23} = \frac{f_2 + 477}{2} \\[1em] \Rightarrow 16f_2 \times 2 = 23(f_2 + 477) \\[1em] \Rightarrow 32f_2 = 23f_2 + 10971 \\[1em] \Rightarrow 32f_2 - 23f_2 = 10971 \\[1em] \Rightarrow 9f_2 = 10971 \\[1em] \Rightarrow f_2 = \frac{10971}{9} \\[1em] \Rightarrow f_2 = 1219.\end{aligned}$$

Substituting value of f₂ in equation (1), we get :

$$\begin{aligned}\Rightarrow f_1 = \frac{16f_2}{23} \\[1em] = \frac{16 \times 1219}{23} \\[1em] = 16 \times 53 \\[1em] = 848.\end{aligned}$$

Hence, original price of fans are ₹ 848 and ₹ 1219.

Given,

y is the mean proportional between x and z

$\therefore \frac{x}{y} = \frac{y}{z} \Rightarrow y^2 = xz.$

To prove,

xy + yz is the mean proportional between x² + y² and y² + z²

$$\begin{aligned}\therefore \frac{x^2 + y^2}{xy + yz} = \frac{xy + yz}{y^2 + z^2} \\[1em] \Rightarrow (xy + yz)(xy + yz) = (x^2 + y^2)(y^2 + z^2) \\[1em] \Rightarrow x^2y^2 + xy^2z + xy^2z + y^2z^2 = x^2y^2 + x^2z^2 + y^4 + y^2z^2 ....[\text{i}]\end{aligned}$$

Substituting y² = xz in L.H.S. of (i)

⇒ x²(xz) + x(xz)z + x(xz)z + (xz)z²

⇒ x³z + x²z² + x²z² + xz³

⇒ x³z + 2x²z² + xz³ .........(ii)

Substituting y² = xz in R.H.S. of (i)

⇒ x²(xz) + x²z² + (xz)² + (xz)z²

⇒ x³z + x²z² + x²z² + xz³

⇒ x³z + 2x²z² + xz³ .........(iii)

Since, (ii) = (iii)

Hence, proved that xy + yz is the mean proportional between x² + y² and y² + z².

Applying componendo and dividendo we get,

$$\begin{aligned}\Rightarrow \frac{\sqrt{3x + 4} + \sqrt{3x - 5} + \sqrt{3x + 4} - \sqrt{3x - 5}}{\sqrt{3x + 4} + \sqrt{3x - 5} - (\sqrt{3x + 4} - \sqrt{3x - 5})} = \frac{9 + 1}{9 - 1} \\[1em] \Rightarrow \frac{2\sqrt{3x + 4}}{2\sqrt{3x - 5}} = \frac{10}{8} \\[1em] \Rightarrow \frac{\sqrt{3x + 4}}{\sqrt{3x - 5}} = \frac{10}{8} \\[1em] \Rightarrow \frac{3x + 4}{3x - 5} = \frac{100}{64} \\[1em]\end{aligned}$$

Again applying componendo and dividendo we get,

$$\begin{aligned}\Rightarrow \frac{3x + 4 + 3x - 5}{3x + 4 - (3x - 5)} = \frac{100 + 64}{100 - 64} \\[1em] \Rightarrow \frac{6x - 1}{9} = \frac{164}{36} \\[1em] \Rightarrow 36(6x - 1) = 1476 \\[1em] \Rightarrow 216x - 36 = 1476 \\[1em] \Rightarrow 216x = 1512 \\[1em] \Rightarrow x = \frac{1512}{216} \\[1em] \Rightarrow x = 7.\end{aligned}$$

Hence, x = 7.

By factor theorem,

If (x - a) is a factor of polynomial f(x), then remainder f(a) = 0.

⇒ x + 2 = 0

⇒ x = -2.

Since, x + 2 is a factor of x³ + ax + b

∴ Remainder = 0.

⇒ (-2)³ + (-2)a + b = 0

⇒ -8 - 2a + b = 0

⇒ b = 2a + 8 ................(1)

x - 3 = 0

⇒ x = 3.

Since, x - 3 is a factor of x³ + ax + b

∴ Remainder = 0.

⇒ (3)³ + (3)a + b = 0

⇒ 27 + 3a + b = 0

⇒ b = -27 - 3a ................(2)

From (1) and (2), we get:

⇒ 2a + 8 = -27 - 3a

⇒ 2a + 3a = -27 - 8

⇒ 5a = -35

⇒ a = $\frac{-35}{5}$

⇒ a = -7.

Substituting value of a in equation (1), we get :

⇒ b = 2a + 8 = 2 × -7 + 8 = -14 + 8 = -6.

f(x) = x³ + ax + b = x³ - 7x - 6.

Substituting x = -1 in f(x), we get :

⇒ (-1)³ - 7(-1) - 6

⇒ -1 + 7 - 6

⇒ -1 + 1

⇒ 0.

∴ (x + 1) is a factor of x³ - 7x - 6.

Hence, a = -7, b = -6 and (x + 1) is the remaining factor.

For x = -2, the value of the given expression

= 2(-2)³ + 9(-2)² + 7(-2) - 6

= 2 × -8 + 9 × 4 - 14 - 6

= -16 + 36 - 14 - 6

= -36 + 36

= 0.

⇒ x + 2 is a factor of 2x³ + 9x² + 7x - 6.

Now dividing 2x³ + 9x² + 7x - 6 by (x + 2),

$$\begin{array}{l} \phantom{x + 2)}{2x^2 + 5x - 3} \\ x + 2\overline{\smash{\big)}2x^3 + 9x^2 + 7x - 6} \\ \phantom{x + 2}\underline{\underset{-}{}2x^3 \underset{-}{+}4x^2} \\ \phantom{{x + 2}2x^3+2}5x^2 + 7x - 6 \\ \phantom{{x + 2}x^3+2}\underline{\underset{-}{}5x^2 \underset{-}{+} 10x} \\ \phantom{{x + 2}x^3+2-5x^2}-3x - 6 \\ \phantom{{x + 2}x^3+2-5x^23}\underline{\underset{+}{-}3x \underset{+}{-} 6} \\ \phantom{{x + 2}x^3+2-5x^23-3}\times \end{array}$$

we get quotient = 2x² + 5x - 3

∴ 2x³ + 9x² + 7x - 6 = (x + 2)(2x² + 5x - 3)

= (x + 2)(2x² + 6x - x - 3)

= (x + 2)[2x(x + 3) - 1(x + 3)]

= (x + 2)(2x - 1)(x + 3).

∴ (x + 2), (2x - 1) and (x + 3) are the factors of 2x³ + 9x² + 7x - 6.

⇒ x + 2 = 0

⇒ x = -2

⇒ 2x - 1 = 0

⇒ x = $\frac{1}{2}$

⇒ x + 3 = 0

⇒ x = -3.

Hence, 2x³ + 9x² + 7x - 6 = (x + 2)(2x - 1)(x + 3) and x = -2, -3, $\frac{1}{2}$.

By remainder theorem,

If f(x), a polynomial in x, is divided by (x - a), the remainder = f(a).

∴ On dividing 2x³ + 5x² - 2x + 8 by (x - a)

Remainder = 2a³ + 5a² - 2a + 8

Given, remainder = 2a³ + 5a².

∴ 2a³ + 5a² - 2a + 8 = 2a³ + 5a²

⇒ -2a + 8 = 0

⇒ 2a = 8

⇒ a = 4.

Hence, a = 4.

Let a be added to x³ - 9x² - 2x + 3 so that the remainder may be 5 when divided by (x - 2).

By remainder theorem,

If f(x), a polynomial in x, is divided by (x - a), the remainder = f(a).

∴ On dividing x³ - 9x² - 2x + 3 + a by (x - 2)

Remainder = (2)³ - 9(2)² - 2(2) + 3 + a

Since, remainder = 5.

∴ 8 - 9(4) - 4 + 3 + a = 5

⇒ 8 - 36 - 1 + a = 5

⇒ a - 29 = 5

⇒ a = 5 + 29 = 34.

Hence, 34 must be added to x³ - 9x² - 2x + 3 so that the remainder may be 5 when divided by (x - 2).

On dividing x³ + 2x² - 5ax - 7 by (x + 1), we get remainder R₁ :

∴ R₁ = (-1)³ + 2(-1)² - 5a(-1) - 7

⇒ R₁ = -1 + 2 + 5a - 7

⇒ R₁ = 5a - 6.

On dividing x³ + ax² - 12x + 6 by (x - 2), we get remainder R₂

∴ R₂ = (2)³ + a(2)² - 12(2) + 6

⇒ R₂ = 8 + 4a - 24 + 6

⇒ R₂ = 4a - 10.

Given,

⇒ 2R₁ + R₂ = 6

⇒ 2(5a - 6) + 4a - 10 = 6

⇒ 10a - 12 + 4a - 10 = 6

⇒ 14a - 22 = 6

⇒ 14a = 28

⇒ a = $\frac{28}{14}$ = 2.

Hence, a = 2.

Let order of matrix B = m × n

Given,

AB = 3C

$$\begin{bmatrix*}[r] 1 & 5 \\ 1 & 2 \end{bmatrix*}_{2 \times 2}B_{m \times n} = 3\begin{bmatrix*}[r] 2 \\ 1 \end{bmatrix*}_{2 \times 1}$$

Since, the product of matrices is possible, only when the number of columns in the first matrix is equal to the number of rows in the second.

∴ m = 2

Also, the no. of columns of product (resulting) matrix is equal to no. of columns of second matrix.

∴ n = 1

Order of matrix B = m × n = 2 × 1.

Let matrix B = $$\begin{bmatrix*}[r] a \\ b \end{bmatrix*}$$

Substituting matrix in AB = 3C we get,

$$\Rightarrow \begin{bmatrix*}[r] 1 & 5 \\ 1 & 2 \end{bmatrix*}\begin{bmatrix*}[r] a \\ b \end{bmatrix*} = 3\begin{bmatrix*}[r] 2 \\ 1 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 1 \times a + 5 \times b \\ 1 \times a + 2 \times b \end{bmatrix*} = \begin{bmatrix*}[r] 6 \\ 3 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] a + 5b \\ a + 2b \end{bmatrix*} = \begin{bmatrix*}[r] 6 \\ 3 \end{bmatrix*}.$$

∴ a + 5b = 6 ........(1)

a + 2b = 3 ........(2)

Subtracting equation (2) from (1), we get :

⇒ a + 5b - (a + 2b) = 6 - 3

⇒ a - a + 5b - 2b = 3

⇒ 3b = 3

⇒ b = 1.

Substituting value of b in (1), we get :

⇒ a + 5(1) = 6

⇒ a + 5 = 6

⇒ a = 1.

B = $$\begin{bmatrix*}[r] a \\ b \end{bmatrix*} = \begin{bmatrix*}[r] 1 \\ 1 \end{bmatrix*}$$

Hence, B = $$\begin{bmatrix*}[r] 1 \\ 1 \end{bmatrix*}$$.

Given,

⇒ 2A + B = $$\begin{bmatrix*}[r] 3 & -4 \\ 2 & 7 \end{bmatrix*}$$ .........(1)

⇒ A - 2B = $$\begin{bmatrix*}[r] 4 & 3 \\ 1 & 1 \end{bmatrix*}$$ .........(2)

Multiplying equation (1) by 2 and adding in equation (2), we get :

$$\Rightarrow 2(2A + B) + A - 2B = 2\begin{bmatrix*}[r] 3 & -4 \\ 2 & 7 \end{bmatrix*} + \begin{bmatrix*}[r] 4 & 3 \\ 1 & 1 \end{bmatrix*} \\[1em] \Rightarrow 4A + 2B + A - 2B = \begin{bmatrix*}[r] 6 & -8 \\ 4 & 14 \end{bmatrix*} + \begin{bmatrix*}[r] 4 & 3 \\ 1 & 1 \end{bmatrix*} \\[1em] \Rightarrow 5A = \begin{bmatrix*}[r] 6 + 4 & -8 + 3 \\ 4 + 1 & 14 + 1 \end{bmatrix*} \\[1em] \Rightarrow 5A = \begin{bmatrix*}[r] 10 & -5 \\ 5 & 15 \end{bmatrix*} \\[1em] \Rightarrow A = \frac{1}{5}\begin{bmatrix*}[r] 10 & -5 \\ 5 & 15 \end{bmatrix*} \\[1em] \Rightarrow A = \begin{bmatrix*}[r] 2 & -1 \\ 1 & 3 \end{bmatrix*}.$$

Substituting value of A in equation (1), we get :

$$\Rightarrow 2\begin{bmatrix*}[r] 2 & -1 \\ 1 & 3 \end{bmatrix*} + B = \begin{bmatrix*}[r] 3 & -4 \\ 2 & 7 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 4 & -2 \\ 2 & 6 \end{bmatrix*} + B = \begin{bmatrix*}[r] 3 & -4 \\ 2 & 7 \end{bmatrix*} \\[1em] \Rightarrow B = \begin{bmatrix*}[r] 3 & -4 \\ 2 & 7 \end{bmatrix*} - \begin{bmatrix*}[r] 4 & -2 \\ 2 & 6 \end{bmatrix*} \\[1em] \Rightarrow B = \begin{bmatrix*}[r] 3 - 4 & -4 - (-2) \\ 2 - 2 & 7 - 6 \end{bmatrix*} \\[1em] \Rightarrow B = \begin{bmatrix*}[r] -1 & -2 \\ 0 & 1 \end{bmatrix*}.$$

Hence, $$A = \begin{bmatrix*}[r] 2 & -1 \\ 1 & 3 \end{bmatrix*}, B = \begin{bmatrix*}[r] -1 & -2 \\ 0 & 1 \end{bmatrix*}.$$

Given,

$$\text{3A - 5B + 2X} = \begin{bmatrix*}[r] 4 & 3 \\ 0 & 1 \end{bmatrix*}$$

Substituting value of A and B in above equation we get,

$$\Rightarrow 3\begin{bmatrix*}[r] 3 & 1 \\ 4 & 0 \end{bmatrix*} - 5\begin{bmatrix*}[r] 1 & -2 \\ 2 & 3 \end{bmatrix*} + 2X = \begin{bmatrix*}[r] 4 & 3 \\ 0 & 1 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 9 & 3 \\ 12 & 0 \end{bmatrix*} - \begin{bmatrix*}[r] 5 & -10 \\ 10 & 15 \end{bmatrix*} + 2X = \begin{bmatrix*}[r] 4 & 3 \\ 0 & 1 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 9 - 5 & 3 - (-10) \\ 12 - 10 & 0 - 15 \end{bmatrix*} + 2X = \begin{bmatrix*}[r] 4 & 3 \\ 0 & 1 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 4 & 13 \\ 2 & -15 \end{bmatrix*} + 2X = \begin{bmatrix*}[r] 4 & 3 \\ 0 & 1 \end{bmatrix*} \\[1em] \Rightarrow 2X = \begin{bmatrix*}[r] 4 & 3 \\ 0 & 1 \end{bmatrix*} - \begin{bmatrix*}[r] 4 & 13 \\ 2 & -15 \end{bmatrix*} \\[1em] \Rightarrow 2X = \begin{bmatrix*}[r] 4 - 4 & 3 - 13 \\ 0 - 2 & 1 - (-15) \end{bmatrix*} \\[1em] \Rightarrow 2X = \begin{bmatrix*}[r] 0 & -10 \\ -2 & 16 \end{bmatrix*} \\[1em] \Rightarrow X = \frac{1}{2}\begin{bmatrix*}[r] 0 & -10 \\ -2 & 16 \end{bmatrix*} \\[1em] \Rightarrow X = \begin{bmatrix*}[r] 0 & -5 \\ -1 & 8 \end{bmatrix*}.$$

Hence, X = $$\begin{bmatrix*}[r] 0 & -5 \\ -1 & 8 \end{bmatrix*}.$$

Let order of matrix X = m × n

Given,

AX = B

$$\begin{bmatrix*}[r] 3 & 4 \\ 4 & -3 \end{bmatrix*}_{2 \times 2}X_{m \times n} = \begin{bmatrix*}[r] 24 \\ 7 \end{bmatrix*}_{2 \times 1}$$

Since, the product of matrices is possible, only when the number of columns in the first matrix is equal to the number of rows in the second.

∴ m = 2

Also, the no. of columns of product (resulting) matrix is equal to no. of columns of second matrix.

∴ n = 1

Order of matrix X = m × n = 2 × 1.

Let matrix X = $$\begin{bmatrix*}[r] a \\ b \end{bmatrix*}$$

Substituting matrix in AX = B we get,

$$\Rightarrow \begin{bmatrix*}[r] 3 & 4 \\ 4 & -3 \end{bmatrix*}\begin{bmatrix*}[r] a \\ b \end{bmatrix*} = \begin{bmatrix*}[r] 24 \\ 7 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 3 \times a + 4 \times b \\ 4 \times a + (-3) \times b \end{bmatrix*} = \begin{bmatrix*}[r] 24 \\ 7 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 3a + 4b \\ 4a - 3b \end{bmatrix*} = \begin{bmatrix*}[r] 24 \\ 7 \end{bmatrix*}.$$

∴ 3a + 4b = 24 ........(1)

4a - 3b = 7 ........(2)

⇒ 3a + 4b = 24

⇒ 3a = 24 - 4b

⇒ a = $\frac{24 - 4b}{3}$ ........(3)

Substituting value of a from equation (3) in equation (2), we get :

$$\begin{aligned}\Rightarrow 4 \times \frac{24 - 4b}{3} - 3b = 7 \\[1em] \Rightarrow \frac{96 - 16b}{3} - 3b = 7 \\[1em] \Rightarrow \frac{96 - 16b - 9b}{3} = 7 \\[1em] \Rightarrow 96 - 25b = 21 \\[1em] \Rightarrow 25b = 96 - 21 \\[1em] \Rightarrow 25b = 75 \\[1em] \Rightarrow b = \frac{75}{25} = 3.\end{aligned}$$

Substituting value of b in equation (3), we get :

a = $\frac{24 - 4b}{3} = \frac{24 - 4 \times 3}{3} = \frac{12}{3}$ = 4.

Hence, X = $$\begin{bmatrix*}[r] 4 \\ 3 \end{bmatrix*}$$

Let order of matrix X = m × n

Given,

XA = B

$$X_{m \times n}\begin{bmatrix*}[r] 3 & 6 \\ -2 & -8 \end{bmatrix*}_{2 \times 2} = \begin{bmatrix*}[r] -2 & 16 \end{bmatrix*}_{1 \times 2}$$

Since, the product of matrices is possible only when the number of columns in the first matrix is equal to the number of rows in the second.

∴ n = 2

Also, the no. of rows of product (resulting) matrix is equal to no. of rows of first matrix.

∴ m = 1

Order of matrix X = m × n = 1 × 2.

Let matrix X = $$\begin{bmatrix*}[r] a & b \end{bmatrix*}$$

Substituting matrix in XA = B we get,

$$\Rightarrow \begin{bmatrix*}[r] a & b \end{bmatrix*}\begin{bmatrix*}[r] 3 & 6 \\ -2 & -8 \end{bmatrix*} = \begin{bmatrix*}[r] -2 & 16 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] a \times 3 + b \times -2 & a \times 6 + b \times -8 \end{bmatrix*} = \begin{bmatrix*}[r] -2 & 16 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 3a - 2b & 6a - 8b \end{bmatrix*} = \begin{bmatrix*}[r] -2 & 16 \end{bmatrix*}$$

∴ 3a - 2b = -2 .........(1)

6a - 8b = 16 .........(2)

⇒ 3a - 2b = -2

⇒ 2(3a - 2b) = 2(-2)

⇒ 6a - 4b = -4 .........(3)

Subtracting equation (2) from (3), we get :

⇒ 6a - 4b - (6a - 8b) = -4 - 16

⇒ 6a - 6a - 4b + 8b = -20

⇒ 4b = -20

⇒ b = $\frac{-20}{4}$

⇒ b = -5.

Substituting value of b in equation (1), we get :

⇒ 3a - 2(-5) = -2

⇒ 3a + 10 = -2

⇒ 3a = -2 - 10

⇒ 3a = -12

⇒ a = $\frac{-12}{3}$

⇒ a = -4.

Hence, X = $$\begin{bmatrix*}[r] -4 & -5 \end{bmatrix*}$$.

We know that,

⇒ a_n = a + (n - 1)d

⇒ a₂ = a + (2 - 1)d

⇒ 11 = a + d

⇒ 11 = a + 9

⇒ a = 11 - 9 = 2.

a₁₅ = a + (15 - 1)d

= 2 + 14 × 9

= 2 + 126

= 128.

Hence, 15th term = 128.

Three digit numbers which are divisible by 7 are 105, 112, ......., 994.

Let there be n terms.

In above series,

a_n = 994, a = 105 and d = 7.

By formula,

⇒ a_n = a + (n - 1)d

⇒ 994 = 105 + (n - 1) × 7

⇒ 994 = 105 + 7(n - 1)

⇒ 994 - 105 = 7(n - 1)

⇒ 889 = 7(n - 1)

⇒ n - 1 = $\frac{889}{7}$

⇒ n - 1 = 127

⇒ n = 128.

Hence, there are 128 three digit numbers, which are divisible by 7.

Let there be n terms in the above series.

From above series,

a_n = 89, a = 4 and d = 5.

By formula,

⇒ a_n = a + (n - 1)d

⇒ 89 = 4 + 5(n - 1)

⇒ 85 = 5(n - 1)

⇒ 17 = n - 1

⇒ n = 18.

By formula,

Sum of A.P. = $\frac{n}{2}(a + a_n)$

Substituting values we get,

$$\begin{aligned}= \frac{18}{2}(4 + 89) \\[1em] = 9 \times 93 \\[1em] = 837.\end{aligned}$$

Hence, sum of A.P. = 837.

2.75, 3.00, 3.25, ..............

The above A.P. has first term (a) = 2.75 and common difference (d) = 3.00 - 2.75 = 0.25

By formula,

n^th term = a_n = a + (n - 1)d.

(i) The amount saved by Daya on 14^th day = 14th term of A.P.

a₁₄ = a + (14 - 1)d

= a + 13d

= 2.75 + 13 × 0.25

= 2.75 + 3.25

= ₹ 6.

Hence, the amount saved by Daya on 14^th day = ₹ 6.

(ii) The amount saved by Daya on 30^th day = 30th term of A.P.

a₃₀ = a + (30 - 1)d

= a + 29d

= 2.75 + 29 × 0.25

= 2.75 + 7.25

= ₹ 10.

Hence, the amount saved by Daya on 30^th day = ₹ 10.

(iii) Sum of A.P. = $\frac{n}{2}[2a + (n - 1)d]$

Total amount saved by Daya in 30 days = Sum of A.P. upto 30 terms.

$$\begin{aligned}= \frac{30}{2}[2 \times 2.75 + (30 - 1) \times 0.25] \\[1em] = 15 \times [5.5 + 29 \times 0.25] \\[1em] = 15 \times [5.5 + 7.25] \\[1em] = 15 \times 12.75 \\[1em] = ₹ 191.25\end{aligned}$$

Hence, total amount saved by Daya in 30 days = ₹ 191.25

Let a be the first term and d be common difference of the A.P.

Given,

Sum of first m terms of an A.P. is n.

⇒ S_m = n

⇒ $\frac{m}{2}[2a + (m - 1)d] = n$

⇒ m[2a + (m - 1)d] = 2n

⇒ 2am + m(m - 1)d = 2n ..........(1)

Sum of first n terms of an A.P. is m.

⇒ S_n = m

⇒ $\frac{n}{2}[2a + (n - 1)d] = m$

⇒ n[2a + (n - 1)d] = 2m

⇒ 2an + n(n - 1)d = 2m ..........(2)

Subtracting eq. (2) from (1), we get

$$\begin{aligned}\Rightarrow 2n - 2m = 2am + m(m - 1)d - [2an + n(n - 1)d] \\[1em] \Rightarrow -2m + 2n = 2am - 2an + m(m - 1)d - n(n - 1)d \\[1em] \Rightarrow -2(m - n) = 2a(m - n) + m^2d - md - n^2d + nd \\[1em] \Rightarrow -2(m - n) = 2a(m - n) + d(m^2 - n^2) - d(m - n) \\[1em] \Rightarrow -2(m - n) = 2a(m - n) + d(m - n)(m + n) - d(m - n) \\[1em] \Rightarrow -2 = 2a + d(m + n) - d \\[1em] \Rightarrow 2a + d(m + n - 1) = -2 .........(3)\end{aligned}$$

S_{m + n} = $\frac{m + n}{2}[a + a + (m + n - 1)d]$

= $\frac{m + n}{2}[2a + (m + n - 1)d]$

= $\frac{m + n}{2} \times -2$ [From (3)]

= -(m + n).

Hence, proved that sum of first (m + n) terms of it is -(m + n).

Let first term of G.P. be a and common ratio be r.

We know that,

n^th term of G.P. = ar^{n - 1}

Given,

3^rd term of a G.P. = 27

∴ ar² = 27 ..........(1)

Given,

6^th term of a G.P. = 729

∴ ar⁵ = 729 ..........(2)

Dividing equation (2) by (1), we get :

$$\begin{aligned}\Rightarrow \frac{ar^5}{ar^2} = \frac{729}{27} \\[1em] \Rightarrow r^3 = 27 \\[1em] \Rightarrow r^3 = 3^3 \\[1em] \Rightarrow r = 3.\end{aligned}$$

Substituting value of r in equation (1), we get :

⇒ a(3)² = 27

⇒ 9a = 27

⇒ a = $\frac{27}{9}$

⇒ a = 3.

7th term = ar⁶

= 3(3)⁶

= 3 × 729

= 2187.

Product of first and seventh term = 3 x 2187

= 6561.

Hence, product of first and seventh term = 6561.

Let a, b, c and d be the 5 geometric means 1 and 27.

∴ 1, a, b, c, d, e and 27 are in G.P.

So, first term (a) = 1 and seventh term = 27.

Let common ratio be r.

∴ ar⁶ = 27

⇒ (1)r⁶ = 27

⇒ r⁶ = 3³

⇒ r⁶ = $(\sqrt{3})^6$

⇒ r = $\sqrt{3}$.

Second term = ar = $1 \times \sqrt{3} = \sqrt{3}$,

Third term = ar² = $1 \times (\sqrt{3})^2 = 3$,

Fourth term = ar³ = $1 \times (\sqrt{3})^3 = 3\sqrt{3}$,

Fifth term = ar⁴ = $1 \times (\sqrt{3})^4 = 9$,

Sixth term = ar⁵ = $1 \times (\sqrt{3})^5 = 9\sqrt{3}$.

Hence, 5 geometric means between 1 and 27 = $\sqrt{3}, 3, 3\sqrt{3}, 9, 9\sqrt{3}$.

The above sequence is a G.P. with first term (a) = 96 and common ratio (r) = $\frac{-48}{96} = -\frac{1}{2}$.

By formula,

Sum of G.P. = $\frac{a(1 - r^n)}{1 - r}$

$$\begin{aligned}\text{Sum upto 10 terms} = \frac{96 \times \Big[1 - \Big(-\frac{1}{2}\Big)^{10}\Big]}{1 - \Big(-\frac{1}{2}\Big)} \\[1em] = \frac{96 \times \Big[1 - \frac{1}{1024}\Big]}{1 + \frac{1}{2}} \\[1em] = \frac{96 \times \frac{1023}{1024}}{\frac{3}{2}} \\[1em] = \frac{12 \times \frac{1023}{128}}{\frac{3}{2}} \\[1em] = \frac{12 \times 1023 \times 2}{128 \times 3} \\[1em] = \frac{2046}{32} \\[1em] = 63\frac{15}{16}.\end{aligned}$$

Hence, sum of sequence = $63\frac{15}{16}.$

(i) Given,

⇒ 4 + 44 + 444 + ...... upto n terms

⇒ 4(1 + 11 + 111 + ....... upto n terms)

⇒ $\frac{4}{9}(9 + 99 + 999 + ......$ upto n terms)

⇒ $\frac{4}{9}[(10 - 1) + (10^2 - 1) + ......$ upto n terms]

⇒ $\frac{4}{9}\Big[(10 + 10^2 ....... + 10^n) - (1 + 1 + ........ \text{n times})\Big]$

Above sequence : 10 + 10² + ......... + 10ⁿ

It is a G.P. with common term (a) = 10 and common ratio (r) = 10.

By formula,

Sum of G.P. = $\frac{a(1 - r^n)}{1 - r}$

⇒ $\frac{4}{9}\Big[\frac{10(10^n - 1)}{10 - 1} - n\Big]$

⇒ $\frac{4}{9}\Big[\frac{10(10^n - 1)}{9} - n\Big]$.

Hence, sum of 4 + 44 + 444 + ...... upto n terms = $\frac{4}{9}\Big[\frac{10(10^n - 1)}{9} - n\Big].$

(ii) Given,

$$\begin{aligned}\Rightarrow 0.7 + 0.77 + 0.777 + ....... \text{ upto n terms } \\[1em] \Rightarrow 7(0.1 + 0.11 + 0.111 + ...... \text{ upto n terms }) \\[1em] \Rightarrow \frac{7}{9}(0.9 + 0.99 + 0.999 + ....... \text{ upto n terms }) \\[1em] \Rightarrow \frac{7}{9}[(1 - 0.1) + (1 - 0.01) + (1 - 0.001) + .......... \text{ upto n terms}] \\[1em] \Rightarrow \frac{7}{9}[1 + 1 + ...... \text{ upto n terms } - (0.1 + 0.01+ 0.001 + ........)] \\[1em] \Rightarrow \frac{7}{9} \Big[n - \Big(\frac{1}{10} + \frac{1}{100} + \frac{1}{1000} + ...... \text{ upto n terms}\Big)\Big] \\[1em]\end{aligned}$$

Above sequence : $\frac{1}{10} + \frac{1}{100} + \frac{1}{1000} + .........$

It is a G.P. with first term (a) = $\frac{1}{10}$ and common ratio (r) = $\frac{1}{10}$.

By formula,

Sum of G.P. = $\frac{a(1 - r^n)}{1 - r}$

$$\begin{aligned}\Rightarrow \frac{7}{9} \Big[n - \frac{\frac{1}{10}\Big(1 - \Big(\frac{1}{10}\Big)^n\Big)}{1 - \frac{1}{10}}\Big] \\[1em] \Rightarrow \frac{7}{9} \Big[n -\frac{\frac{1}{10}\Big(1 - \frac{1}{10^n}\Big)}{\frac{9}{10}}\Big] \\[1em] \Rightarrow \frac{7}{9} \Big[n -\frac{\Big(1 - \frac{1}{10^n}\Big)}{9}\Big] \\[1em] \Rightarrow \frac{7}{9} \Big[n - \frac{1}{9}\Big(1 - \frac{1}{10^n}\Big)\Big]\end{aligned}$$

Hence, sum of 0.7 + 0.77 + 0.777 + ....... upto n terms = $\frac{7}{9} \Big[n - \frac{1}{9}\Big(1 - \frac{1}{10^n}\Big)\Big]$.

Let, x = $0.4\overline{23}$.

⇒ x = 0.4232323......

⇒ 10x = 4.232323.......

⇒ 1000x = 423.2323.......

⇒ 1000x - 10x = 423.2323....... - 4.232323.......

⇒ 990x = 419

⇒ x = $\frac{419}{990}$

Hence, $0.4\bar{23} = \frac{419}{990}$.

Let first term of G.P. be A and common ratio be R.

p^th term = a

AR^{p - 1} = a.

q^th term = b

AR^{q - 1} = b.

r^th term = c

AR^{r - 1} = c.

$$\begin{aligned}\Rightarrow a^{q - r}.b^{r - p}.c^{p - q} = (AR^{p - 1})^{q - r}.(AR^{q - 1})^{r - p}.(AR^{r - 1})^{p - q} \\[1em] = A^{q - r}.R^{(p - 1)(q - r)}.A^{r - p}.R^{(q - 1)(r - p)}.A^{p - q}.R^{(r - 1)(p - q)} \\[1em] = A^{q - r + r - p + p - q}.R^{(p - 1)(q - r) + (q - 1)(r - p) + (r - 1)(p - q)} \\[1em] = A^0.R^{(pq - pr - q + r) + (qr - qp - r + p) + (rp - rq - p + q)} \\[1em] = A^0.R^0 \\[1em] = 1.\end{aligned}$$

Hence, proved that

$$\begin{aligned}\Rightarrow a^{q - r}.b^{r - p}.c^{p - q} = 1 \\[1em] \text{Taking log on both sides} \\[1em] \Rightarrow log(a^{q - r}.b^{r - p}.c^{p - q}) = log \space 1 \\[1em] \Rightarrow log(a^{q - r}) + log(b^{r - p}) + log(c^{p - q}) = 0 \\[1em] \Rightarrow (q - r)log \space a + (r - p)log \space b + (p - q)log \space c = 0.\end{aligned}$$

Hence, proved that (q - r)log a + (r - p)log b + (p - q)log c = 0.

Given, a, b and c are in G.P.

∴ b² = ac .......(i)

For, log a, log b and log c to be in A.P,

log b- log a = log c - log b

log b + log b = log a + log c

2log b = log a + log c

L.H.S. = 2log b = log b²

R.H.S. = log ac

Since, b² = ac,

L.H.S. = R.H.S.

∴ 2log b = log a + log c

Hence, proved that log a, log b and log c are in A.P.

Let a, b and c be in G.P.

∴ b² = ac

Let, b² = ac = m ........(i)

Each term be raised to power x.

So, terms = a^x, b^x and c^x.

Terms are in G.P. if common ratio between terms are equal,

$$\begin{aligned}\Rightarrow \frac{b^x}{a^x} = \frac{c^x}{b^x} \\[1em] \Rightarrow b^x.b^x = c^x.a^x \\[1em] \Rightarrow (b^x)^2 = (a^x.c^x) \\[1em] \Rightarrow (b^2)^x = (ac)^x \\[1em] \text{From (i)},\\[1em] \Rightarrow m^x = m^x.\end{aligned}$$

Since, L.H.S. = R.H.S.,

Hence, proved that on raising each term of a G.P. to the power x, the resulting sequence is also a G.P.

Given,

a, b, c are in A.P.

⇒ 2b = a + c ........(i)

a, x, b are in G.P.

⇒ x² = ab ........(ii)

b, y and c are in G.P.

⇒ y² = bc .........(iii)

Adding (ii) and (iii) we get,

⇒ x² + y² = ab + bc

⇒ x² + y² = b(a + c)

⇒ x² + y² = b.2b (From i)

⇒ x² + y² = 2b².

Hence, proved that x², b², y² are in A.P.

Given, x, y and z are in G.P.

$\therefore \frac{y}{x} = \frac{z}{y}$

∴ y² = xz = m (let) .....(1)

New terms = x⁴, y⁴ and z⁴.

For the terms to be in G.P., common ratio between terms must be equal,

$$\begin{aligned}\Rightarrow \frac{y^4}{x^4} = \frac{z^4}{y^4} \\[1em] \Rightarrow (y^4)^2 = x^4.z^4 \\[1em] \Rightarrow (y^2)^4 = (xz)^4\end{aligned}$$

Substituting value from equation (1), we get :

⇒ m⁴ = m⁴.

Since, L.H.S. = R.H.S.

Hence, proved that x⁴, y⁴ and z⁴ are in G.P.

Common ratio (r) = $\frac{\sqrt{15}}{\sqrt{5}} = \frac{\sqrt{5} \times \sqrt{3}}{\sqrt{5}} = \sqrt{3}$.

By formula,

Sum of first n terms of G.P. = $\frac{a(r^n - 1)}{(r - 1)}$

Substituting values we get :

$(S) = \frac{\sqrt{5}[(\sqrt{3})^n - 1]}{\sqrt{3} - 1}$.

Hence, sum of first n terms of the G.P. $\sqrt{5}, \sqrt{15}, \sqrt{45} ..... \text{ is } \frac{\sqrt{5}[(\sqrt{3})^n - 1]}{\sqrt{3} - 1}.$

(i) We know that,

On reflection in x-axis, the sign of y-coordinate changes.

∴ (4, -3) on reflection in x-axis gives (4, 3).

Given,

(4, -3) on reflection in x-axis gives (-m, n)

∴ (-m, n) = (4, 3)

⇒ -m = 4 and n = 3

⇒ m = -4 and n = 3.

Hence, m = -4 and n = 3.

(ii) We know that,

On reflection in y-axis, the sign of x-coordinate changes.

∴ (m, 5) on reflection in y-axis gives (-m, 5).

Given,

(m, 5) on reflection in y-axis gives (-5, n - 2)

∴ (-m, 5) = (-5, n - 2)

⇒ -m = -5 and n - 2 = 5

⇒ m = 5 and n = 7.

Hence, m = 5 and n = 7.

(iii) We know that,

On reflection in origin, the sign of x-coordinate and y-coordinate changes.

∴ (-6, n + 2) on reflection in origin gives [6, -(n + 2)].

Given,

(-6, n + 2) on reflection in origin gives (m + 3, -4)

∴ [6, -(n + 2)] = (m + 3, -4)

⇒ 6 = m + 3 and -(n + 2) = -4

⇒ m = 6 - 3 and -n - 2 = -4

⇒ m = 3 and -n = -4 + 2

⇒ m = 3 and -n = -2

⇒ m = 3 nd n = 2.

Hence, m = 3 and n = 2.

(i) x = 0 is the y-axis.

On reflection in y-axis,

A(-2, 4) ⇒ A'(2, 4).

Hence, coordinates of A' = (2, 4).

(ii) On reflection in y-axis,

B(-4, 1) ⇒ B'(4, 1).

Hence, coordinates of B' = (4, 1).

(iii) From graph,

On reflection in line BB'

A(-2, 4) ⇒ A"(-2, -2).

Hence, coordinates of A" = (-2, -2).

From graph,

Angle between lines A'A" and BB' = 56° 19' and B'A'BA" is a parallelogram.

(i) From graph,

Coordinates of A = (-4, 5) and B = (7, 0).

(ii) By distance formula,

OA = $\sqrt{(-4 - 0)^2 + (5 - 0)^2} = \sqrt{(-4)^2 + 5^2} = \sqrt{16 + 25} = \sqrt{41}$ units.

OA₁ = $\sqrt{(4 - 0)^2 + (-5 - 0)^2} = \sqrt{16 + 25} = \sqrt{41}$ units.

OB = OB₁ = 7 units.

Since, diagonals bisect each other.

Hence, ABA₁B₁ is a parallelogram.

(iii) We know that,

On reflection in x-axis, the sign of y-coordinate changes.

∴ (-4, 5) on reflection in x-axis gives (-4, -5).

We know that,

On reflection in y-axis, the sign of x-coordinate changes.

∴ (-4, -5) on reflection in y-axis gives (4, -5).

∴ A₂ = (4, -5).

Hence, coordinates of A₂ = (4, -5).

(iv) We know that,

On reflection in y-axis, the sign of x-coordinate changes.

∴ (7, 0) on reflection in y-axis gives (-7, 0).

We know that,

On reflection in x-axis, the sign of y-coordinate changes.

∴ (-7, 0) on reflection in x-axis gives (-7, 0).

∴ B₂ = (-7, 0).

Hence, coordinates of B₂ = (-7, 0).

Let the point M(p, -1) divide the line segment joining the points A(1, -3) and B(6, 2) in ratio k : 1.

By section formula,

$y = \frac{m_1y_2 + m_2y_1}{m_1 + m_2}$

Substituting value we get,

$$\begin{aligned}\Rightarrow -1 = \frac{k \times 2 + 1 \times -3}{k + 1} \\[1em] \Rightarrow -1 = \frac{2k - 3}{k + 1} \\[1em] \Rightarrow -(k + 1) = 2k - 3 \\[1em] \Rightarrow -k - 1 = 2k - 3 \\[1em] \Rightarrow 2k + k = -1 + 3 \\[1em] \Rightarrow 3k = 2 \\[1em] \Rightarrow k = \frac{2}{3}.\end{aligned}$$

⇒ k : 1 = $\frac{2}{3} : 1$ = 2 : 3.

By section formula,

$x = \frac{m_1x_2 + m_2x_1}{m_1 + m_2}$

Substituting value we get,

$$\begin{aligned}\Rightarrow p = \frac{2 \times 6 + 3 \times 1}{2 + 3} \\[1em] \Rightarrow p = \frac{12 + 3}{5} \\[1em] \Rightarrow p = \frac{15}{5} \\[1em] \Rightarrow p = 3.\end{aligned}$$

Hence, ratio = 2 : 3 and p = 3.

By formula,

Centroid of triangle = $\Big(\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}\Big)$

Given,

Centroid of triangle = (0, 0)

Substituting values for x-coordinate we get,

$$\begin{aligned}\Rightarrow 0 = \Big(\frac{-4 + x + 6}{3}\Big) \\[1em] \Rightarrow -4 + x + 6 = 0 \\[1em] \Rightarrow x + 2 = 0 \\[1em] \Rightarrow x = -2.\end{aligned}$$

Substituting values for y-coordinate we get,

$$\begin{aligned}\Rightarrow 0 = \Big(\frac{4 + (-1) + y}{3}\Big) \\[1em] \Rightarrow 4 - 1 + y = 0 \\[1em] \Rightarrow y + 3 = 0 \\[1em] \Rightarrow y = -3.\end{aligned}$$

Hence, x = -2 and y = -3.

(a) Let coordinates of P be (x, y)

By section formula,

P(x, y) = $\Big(\frac{m_1x_2 + m_2x_1}{m_1 + m_2}, \frac{m_1y_2 + m_2y_1}{m_1 + m_2}\Big)$

Substituting values we get,

$$\begin{aligned}\Rightarrow P = \Big(\frac{1 \times -1 + 2 \times 2}{1 + 2}, \frac{1 \times 2 + 2 \times 5}{1 + 2}\Big) \\[1em] \Rightarrow P = \Big(\frac{-1 + 4}{3}, \frac{2 + 10}{3}\Big) \\[1em] \Rightarrow P = \Big(\frac{3}{3}, \frac{12}{3}\Big) \\[1em] \Rightarrow P = (1, 4).\end{aligned}$$

Let coordinates of Q be (a, b)

By section formula,

Q(a, b) = $\Big(\frac{m_1x_2 + m_2x_1}{m_1 + m_2}, \frac{m_1y_2 + m_2y_1}{m_1 + m_2}\Big)$

Substituting values we get,

$$\begin{aligned}\Rightarrow Q = \Big(\frac{1 \times 5 + 2 \times 2}{1 + 2}, \frac{1 \times 8 + 2 \times 5}{1 + 2}\Big) \\[1em] \Rightarrow Q = \Big(\frac{5 + 4}{3}, \frac{8 + 10}{3}\Big) \\[1em] \Rightarrow Q = \Big(\frac{9}{3}, \frac{18}{3}\Big) \\[1em] \Rightarrow Q = (3, 6).\end{aligned}$$

Hence, coordinates of P = (1, 4) and Q = (3, 6).

(b) By distance formula,

$$\begin{aligned}BC = \sqrt{[5 - (-1)]^2 + (8 - 2)^2} \\[1em] = \sqrt{[5 + 1]^2 + 6^2} \\[1em] = \sqrt{6^2 + 6^2} \\[1em] = \sqrt{36 + 36} \\[1em] = \sqrt{72} = 6\sqrt{2} \text{ units}. \\[1em] PQ = \sqrt{(3 - 1)^2 + (6 - 4)^2} \\[1em] = \sqrt{2^2 + 2^2} \\[1em] = \sqrt{4 + 4} \\[1em] = \sqrt{8} \\[1em] = 2\sqrt{2}.\end{aligned}$$

3PQ = 3 × $2\sqrt{2} = 6\sqrt{2}$ = BC.

Hence, proved that BC = 3PQ.

Let the points be A(a, b), B(a + 3, b + 4), C(a - 1, b + 7) and D(a - 4, b + 3).

We know that,

Distance formula = $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

By distance formula,

$$\begin{aligned}AB = \sqrt{(a + 3 - a)^2 + (b + 4 - b)^2} \\[1em] = \sqrt{3^2 + 4^2} \\[1em] = \sqrt{9 + 16} \\[1em] = \sqrt{25} \\[1em] = 5\text{ units}. \\[1em] BC = \sqrt{[(a - 1) - (a + 3)]^2 + [(b + 7) - (b + 4)]^2} \\[1em] =\sqrt{[a - a - 1 - 3]^2 + [b - b + 7 - 4]^2} \\[1em] = \sqrt{[-4]^2 + [3]^2} \\[1em] = \sqrt{16 + 9} \\[1em] = \sqrt{25} \\[1em] = 5 \text{ units}. \\[1em] CD = \sqrt{[(a - 4) - (a - 1)]^2 + [(b + 3) - (b + 7)]^2} \\[1em] = \sqrt{[a - a - 4 + 1]^2 + [b - b + 3 - 7]^2} \\[1em] = \sqrt{[-3]^2 + [-4]^2} \\[1em] = \sqrt{9 + 16} \\[1em] = \sqrt{25} \\[1em] = 5 \text{ units}. \\[1em] DA = \sqrt{[a - (a - 4)]^2 + [b - (b + 3)]^2} \\[1em] = \sqrt{4^2 + (-3)^2} \\[1em] = \sqrt{16 + 9} \\[1em] = \sqrt{25} \\[1em] = 5 \text{ units}.\end{aligned}$$

Since, AB = BC = CD = DA.

Hence, proved that ABCD are the vertices of a parallelogram.

By formula,

Mid-point (M) = $\Big(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\Big)$

Given,

P is mid-point of BC.

$$\begin{aligned}\therefore P = \Big(\frac{0 + (-2)}{2}, \frac{2 + (-4)}{2}\Big) \\[1em] = \Big(\frac{-2}{2}, \frac{-2}{2}\Big) \\[1em] = (-1, -1).\end{aligned}$$

Q is mid-point of CA.

$$\begin{aligned}\therefore Q = \Big(\frac{-2 + (-4)}{2}, \frac{-4 + 2}{2}\Big) \\[1em] = \Big(\frac{-6}{2}, \frac{-2}{2}\Big) \\[1em] = (-3, -1).\end{aligned}$$

R is mid-point of AB.

$$\begin{aligned}\therefore R = \Big(\frac{-4 + 0}{2}, \frac{2 + 2}{2}\Big) \\[1em] = \Big(\frac{-4}{2}, \frac{4}{2}\Big) \\[1em] = (-2, 2).\end{aligned}$$

Centroid of the triangle is given by (G) = $\Big(\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}\Big)$

Let G₁ and G₂ be centroid of △ABC and △PQR.

Substituting values we get,

$$\begin{aligned}G_1 = \Big(\frac{-4 + 0 + (-2)}{3}, \frac{2 + 2 + (-4)}{3}\Big) \\[1em] = \Big(-\frac{6}{3}, \frac{0}{3}\Big) \\[1em] = (-2, 0). \\[1em] G_2 = \Big(\frac{(-1) + (-3) + (-2)}{3}, \frac{(-1) + (-1) + 2}{3}\Big) \\[1em] = \Big(\frac{-6}{3}, \frac{0}{3}\Big) \\[1em] = (-2, 0).\end{aligned}$$

Since, G₁ = G₂.

Hence, proved that the centroid of △PQR is the same as the centroid of △ABC.

(i) By formula,

Mid-point (M) = $\Big(\frac{m_1x_2 + m_2x_1}{m_1 + m_2}, \frac{m_1y_2 + m_2y_1}{m_1 + m_2}\Big)$

Let D be the mid-point of AC.

D = $\Big(\frac{1 + 15}{2}, \frac{5 + 9}{2}\Big) = \Big(\frac{16}{2}, \frac{14}{2}\Big)$ = (8, 7).

By two-point formula,

Equation of straight line :

⇒ y - y₁ = $\frac{y_2 - y_1}{x_2 - x_1}$(x - x₁)

Equation of BD :

$$\begin{aligned}\Rightarrow y - 7 = \frac{7 - 7}{8 - (-3)}[x - (-3)] \\[1em] \Rightarrow y - 7 = 0 \times (x + 3) \\[1em] \Rightarrow y - 7 = 0.\end{aligned}$$

Hence, equation of a line passing through the mid-point of AC and the point B is y - 7 = 0.

(ii) By formula,

Slope = $\frac{y_2 - y_1}{x_2 - x_1}$

Slope of AB = $\frac{7 - 5}{-3 - 1} = -\frac{2}{4} = -\frac{1}{2}$.

As, slope of parallel lines are equal.

∴ Slope of line parallel to AB = -$\frac{1}{2}$.

By point-slope form, equation :

$y - y_1 = m(x - x_1)$

Slope of line through C and parallel to AB.

$$\begin{aligned}\Rightarrow y - 9 = -\frac{1}{2}(x - 15) \\[1em] \Rightarrow 2(y - 9) = -1(x - 15) \\[1em] \Rightarrow 2y - 18 = -x + 15 \\[1em] \Rightarrow x + 2y = 15 + 18 \\[1em] \Rightarrow x + 2y = 33.\end{aligned}$$

Hence, equation of the line through C and parallel to AB is x + 2y = 33.

(iii) Equations are :

⇒ y - 7 = 0 ..........(1)

⇒ x + 2y = 33 .........(2)

From equation (1),

⇒ y = 7 ........(3)

Substituting above value of y in equation (2), we get :

⇒ x + 2(7) = 33

⇒ x + 14 = 33

⇒ x = 33 - 14 = 19.

∴ P = (x, y) = (19, 7).

Hence, P = (19, 7).

(iv) Steps :

1. Mark points A, B, C and P.

2. Join AB, BC, CP and PA.

3. From graph, PABC is a trapezium.

Hence, PABC is a trapezium.

Given, equation of line,

⇒ x - 4y = 6

⇒ x - 4y - 6 = 0

⇒ 4y = x - 6

⇒ y = $\frac{1}{4}x - \frac{6}{4}$.

Comparing with y = mx + c, we get :

Slope = $\frac{1}{4}$.

Since, given line and AB are perpendicular to each other, so their products = -1. Let slope of line AB be m₁.

$\therefore \frac{1}{4} \times m_1 = -1$

⇒ m₁ = -4.

Now, equation of AB can be found by point slope form,

⇒ y - y₁ = m(x - x₁)

⇒ y - 3 = -4(x - 1)

⇒ y - 3 = -4x + 4

⇒ 4x + y - 3 - 4 = 0

⇒ 4x + y - 7 = 0.

Since, line x - 4y - 6 = 0 is perpendicular bisector of 4x + y - 7 = 0 hence solving them simultaneously to find point of intersection,

⇒ x - 4y = 6 .........(1)

⇒ 4x + y = 7 ........(2)

Multiplying (2) with 4 and adding with (1), we get :

⇒ 4(4x + y) + x - 4y = 4 × 7 + 6

⇒ 16x + 4y + x - 4y = 28 + 6

⇒ 17x = 34

⇒ x = $\frac{34}{17}$

⇒ x = 2.

Putting value of x in (1),

⇒ 2 - 4y = 6

⇒ -4y = 4

⇒ y = -1.

Mid-point of AB = (2, -1). Let A be (a, b).

By mid-point formula, coordinates of mid-point of AB are $\Big(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\Big)$

$$\begin{aligned}\therefore (2, -1) = \Big(\frac{a + 1}{2}, \frac{b + 3}{2}\Big) \\[1em] \Rightarrow \frac{a + 1}{2} = 2 \text{ and } \frac{b + 3}{2} = -1 \\[1em] \Rightarrow a + 1 = 4 \text{ and } b + 3 = -2 \\[1em] \Rightarrow a = 3 \text{ and } b = -5.\end{aligned}$$

Hence, coordinates of A = (3, -5).

By two-point formula,

Equation of line : $y - y_1 = \frac{y_2 - y_1}{x_2 - x_1}(x - x_1)$

∴ Equation of line passing through (7, -3) and (2, -2) is

$$\begin{aligned}\Rightarrow y - (-3) = \frac{-2 - (-3)}{2 - 7} \times (x - 7) \\[1em] \Rightarrow y + 3 = \frac{-2 + 3}{-5} \times (x - 7) \\[1em] \Rightarrow y + 3 = -\frac{1}{5}(x - 7) \\[1em] \Rightarrow 5(y + 3) = -(x - 7) \\[1em] \Rightarrow 5y + 15 = -x + 7 \\[1em] \Rightarrow x + 5y + 15 - 7 = 0 \\[1em] \Rightarrow x + 5y + 8 = 0.\end{aligned}$$

Given,

Line meets x-axis at point P. We know that, y-coordinate = 0 at x-axis.

Substituting, y = 0 in x + 5y + 8 = 0, we get :

⇒ x + 5(0) + 8 = 0

⇒ x + 8 = 0

⇒ x = -8.

∴ P = (-8, 0).

Line meets y-axis at point Q. We know that, x-coordinate = 0 at y-axis.

Substituting, x = 0 in x + 5y + 8 = 0, we get :

⇒ 0 + 5y + 8 = 0

⇒ 5y + 8 = 0

⇒ y = -$\frac{8}{5}$.

∴ Q = $\Big(0, -\frac{8}{5}\Big)$.

Hence, equation of line is x + 5y + 8 = 0, P = (-8, 0) and Q = $\Big(0, -\frac{8}{5}\Big)$.

(i) D is the mid-point of AC [∵ BD is median]

By mid-point formula,

$$\begin{aligned}\Rightarrow D = \Big(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\Big) \\[1em] \Rightarrow D = \Big(\frac{-3 + 1}{2}, \frac{1 + (-2)}{2}\Big) \\[1em] \Rightarrow D = \Big(\frac{-2}{2}, \frac{-1}{2}\Big) \\[1em] \Rightarrow D = \Big(-1, -\frac{1}{2}\Big)\end{aligned}$$

By two-point formula,

Equation of line : $y - y_1 = \frac{y_2 - y_1}{x_2 - x_1}(x - x_1)$

∴ Equation of line passing through (4, 4) and $\Big(-1, -\frac{1}{2}\Big)$.

$$\begin{aligned}\Rightarrow y - 4 = \frac{-\frac{1}{2} - 4}{-1 - 4} \times (x - 4) \\[1em] \Rightarrow y - 4 = \frac{\frac{-1 - 8}{2}}{-5} \times (x - 4) \\[1em] \Rightarrow y - 4 = \frac{-\frac{9}{2}}{-5} \times (x - 4) \\[1em] \Rightarrow y - 4 = \frac{9}{10}(x - 4) \\[1em] \Rightarrow 10(y - 4) = 9(x - 4) \\[1em] \Rightarrow 10y - 40 = 9x - 36 \\[1em] \Rightarrow 9x - 10y - 36 + 40 = 0 \\[1em] \Rightarrow 9x - 10y + 4 = 0 \\[1em] \Rightarrow 9x + 4 = 10y.\end{aligned}$$

Hence, equation of median BD is 10y = 9x + 4.

(ii) By formula,

Slope = $\frac{y_2 - y_1}{x_2- x_1}$

Slope of BC = $\frac{-2 - 4}{1 - 4} = \frac{-6}{-3}$ = 2.

We know that,

Product of slope of perpendicular lines = -1.

⇒ Slope of BC × Slope of AE = -1

⇒ 2 × Slope of AE = -1

⇒ Slope of AE = -$\frac{1}{2}$.

By point slope form, equation of AE :

$$\begin{aligned}\Rightarrow y - y_1 = m(x - x_1) \\[1em] \Rightarrow y - 1 = -\frac{1}{2}[x - (-3)] \\[1em] \Rightarrow 2(y - 1) = -[x + 3] \\[1em] \Rightarrow 2y - 2 = -x - 3 \\[1em] \Rightarrow 2y + x = -3 + 2 \\[1em] \Rightarrow x + 2y = -1 \\[1em] \Rightarrow x + 2y + 1 = 0.\end{aligned}$$

Hence, equation of altitude AE is x + 2y + 1 = 0.

Mid-point of line segment joining the points (4, -3) and (3, 1) is $\Big(\frac{4 + 3}{2}, \frac{-3 + 1}{2}\Big) = \Big(\frac{7}{2}, \frac{-2}{2}\Big) = \Big(\frac{7}{2}, -1\Big).$

Slope of line joining the points (4, -3) and (3, 1) is :

$$\begin{aligned}= \frac{y_2 - y_1}{x_2 - x_1} \\[1em] = \frac{1 - (-3)}{3 - 4} \\[1em] = \frac{4}{-1} \\[1em] = -4.\end{aligned}$$

We know that,

⇒ Product of slopes of perpendicular line = -1.

⇒ Slope of line joining the points (4, -3) and (3, 1) × Slope of perpendicular line = -1

⇒ -4 × Slope of perpendicular line = -1

⇒ Slope of perpendicular line = $\frac{1}{4}$.

Perpendicular bisector of the line segment joining the points (4, -3) and (3, 1) will have slope = $\frac{1}{4}$ and will pass through the point $\Big(\frac{7}{2}, -1\Big)$.

By point-slope form, equation of perpendicular bisector is

$$\begin{aligned}\Rightarrow y - y_1 = m(x - x_1) \\[1em] \Rightarrow y - (-1) = \frac{1}{4}\Big(x - \frac{7}{2}\Big) \\[1em] \Rightarrow 4(y + 1) = x - \frac{7}{2} \\[1em] \Rightarrow 4y + 4 = x - \frac{7}{2} \\[1em] \Rightarrow 2(4y + 4) = 2\Big(x - \frac{7}{2}\Big) \\[1em] \Rightarrow 8y + 8 = 2x - 7 \\[1em] \Rightarrow 2x - 8y - 7 - 8 = 0 \\[1em] \Rightarrow 2x - 8y - 15 = 0.\end{aligned}$$

Hence, equation of perpendicular bisector of the line segment joining the points (4, -3) and (3, 1) is 2x - 8y - 15 = 0.

Given 1st equation,

⇒ (p + 1)x + y = 3

⇒ y = -(p + 1)x + 3

Comparing above equation with y = mx + c, we get :

⇒ Slope (m₁) = -(p + 1)

Given 2nd equation,

⇒ 3y - (p - 1)x = 4

⇒ 3y = (p - 1)x + 4

⇒ y = $\frac{(p - 1)x}{3} + \frac{4}{3}$

Comparing above equation with y = mx + c, we get :

⇒ Slope (m₂) = $\frac{p - 1}{3}$

We know that,

Product of slopes of perpendicular lines = -1.

$$\begin{aligned}\therefore -(p + 1) \times \frac{p - 1}{3} = -1 \\[1em] \Rightarrow \frac{-(p^2 - 1)}{3} = -1 \\[1em] \Rightarrow -(p^2 - 1) = -3 \\[1em] \Rightarrow p^2 - 1 = 3 \\[1em] \Rightarrow p^2 = 4 \\[1em] \Rightarrow p = \sqrt{4} = \pm 2.\end{aligned}$$

Hence, p = $\pm 2$.

Given,

⇒ y + (2p + 1)x + 3 = 0

⇒ y = -(2p + 1)x - 3

Comparing above equation with y = mx + c, we get :

⇒ Slope (m₁) = -(2p + 1)

Given,

⇒ 8y - (2p - 1)x = 5

⇒ 8y = (2p - 1)x + 5

⇒ y = $\frac{2p - 1}{8} + \frac{5}{8}$

Comparing above equation with y = mx + c, we get :

⇒ Slope (m₂) = $\frac{2p - 1}{8}$.

We know that,

Product of slopes of perpendicular lines = -1.

$$\begin{aligned}\therefore -(2p + 1) \times \frac{2p - 1}{8} = -1 \\[1em] \Rightarrow \frac{-(4p^2 - 1)}{8} = -1 \\[1em] \Rightarrow -(4p^2 - 1) = -8 \\[1em] \Rightarrow 4p^2 - 1 = 8 \\[1em] \Rightarrow 4p^2 = 9 \\[1em] \Rightarrow p^2 = \frac{9}{4} \\[1em] \Rightarrow p = \sqrt{\frac{9}{4}} = \pm \frac{3}{2} = \pm 1\frac{1}{2}.\end{aligned}$$

Hence, p = $\pm 1\frac{1}{2}$.

The square is shown in the fig below:

Given,

Equation of diagonal BD is x + 2y = 10

⇒ 2y = -x + 10

⇒ y = $-\frac{1}{2}x + \frac{10}{2}$

Comparing it with y = mx + c, we get :

Slope (m₁) = $-\frac{1}{2}$.

We know that,

Diagonals of a square are perpendicular to each other and product of slopes of perpendicular lines = -1.

Let slope of diagonal AC = m₂.

$$\begin{aligned}\therefore m_2 \times -\frac{1}{2} = -1 \\[1em] \Rightarrow -m_2 = -2 \\[1em] \Rightarrow m_2 = 2.\end{aligned}$$

By point-slope form,

Equation of AC is y - y₁ = m(x - x₁)

⇒ y - 2 = 2(x - 1)

⇒ y - 2 = 2x - 2

⇒ 2x - 2 + 2 = y

⇒ y = 2x.

Since, center of the square is the point of intersection of its diagonals.

Solving,

x + 2y = 10 and y = 2x simultaneously, we get :

⇒ x + 2y = 10 .......(1)

⇒ y = 2x .........(2)

Substituting value of y from eq (2) in (1), we get :

⇒ x + 2(2x) = 10

⇒ x + 4x = 10

⇒ 5x = 10

⇒ x = $\frac{10}{5}$

⇒ x = 2.

⇒ y = 2x = 2(2) = 4.

Center = (2, 4).

Hence, equation of other diagonal is y = 2x and coordinates of circle = (2, 4).

Since, points A, B and C are collinear.

∴ Slope of AB = Slope of BC

$$\begin{aligned}\Rightarrow \frac{1- 3}{2 - a} = \frac{a - 1}{5 - 2} \\[1em] \Rightarrow \frac{-2}{2 - a} = \frac{a - 1}{3} \\[1em] \Rightarrow -2 \times 3 = (a - 1)(2 - a) \\[1em] \Rightarrow 2a - a^2 - 2 + a = -6 \\[1em] \Rightarrow 3a - a^2 = -6 + 2 \\[1em] \Rightarrow a^2 - 3a - 4 = 0 \\[1em] \Rightarrow a^2 - 4a + a - 4 = 0 \\[1em] \Rightarrow a(a - 4) + 1(a - 4) = 0 \\[1em] \Rightarrow (a + 1)(a - 4) = 0 \\[1em] \Rightarrow a + 1 = 0 \text{ or } a - 4 = 0 \\[1em] \Rightarrow a = -1 \text{ or } a = 4.\end{aligned}$$

Let a = -1.

A = (-1, 3) and B = (2, 1).

$$\begin{aligned}\text{Slope of AB } = \frac{y_2 - y_1}{x_2 - x_1} \\[1em] = \frac{1 - 3}{2 - (-1)} \\[1em] = \frac{-2}{3}.\end{aligned}$$

By point-slope form,

Equation of AB,

⇒ y - y₁ = m(x - x₁)

⇒ y - 3 = $-\frac{2}{3}$[x - (-1)]

⇒ 3(y - 3) = -2(x + 1)

⇒ 3y - 9 = -2x - 2

⇒ 3y - 9 = -2x - 2

⇒ 2x + 3y = -2 + 9

⇒ 2x + 3y = 7.

Let a = 4.

A = (4, 3) and B = (2, 1).

$$\begin{aligned}\text{Slope of AB } = \frac{y_2 - y_1}{x_2 - x_1} \\[1em] = \frac{1 - 3}{2 - 4} \\[1em] = \frac{-2}{-2} = 1.\end{aligned}$$

By point-slope form,

Equation of AB,

⇒ y - y₁ = m(x - x₁)

⇒ y - 3 = 1[x - 4]

⇒ y - 3 = x - 4

⇒ x - y = -3 + 4

⇒ x - y = 1.

Hence, a = -1 or 4 and equation of line = 2x + 3y = 7 or x - y = 1.

Let A be the point where the line $\frac{x}{a} - \frac{y}{b}$ = 1 meets y-axis.

So, x co-ordinate of point A will be zero.

Substituting x = 0 in equation we get,

$$\begin{aligned}\Rightarrow \frac{0}{a} - \frac{y}{b} = 1 \\[1em] \Rightarrow -\frac{y}{b} = 1 \\[1em] \Rightarrow y = -b.\end{aligned}$$

A = (0, -b).

The given line equation is,

$$\begin{aligned}\Rightarrow \frac{x}{a} - \frac{y}{b} = 1 \\[1em] \Rightarrow \frac{y}{b} = \frac{x}{a} - 1 \\[1em] \Rightarrow y = \frac{b}{a}x - 1.\end{aligned}$$

Comparing above equation with y = mx + c we get,

Slope (m) = $\frac{b}{a}$

Let slope of perpendicular line be m₁.

As product of slope of perpendicular lines is -1,

∴ m × m₁ = -1

⇒ $\frac{b}{a} \times m_1 = -1$

⇒ m₁ = $-\frac{a}{b}$.

Equation of line through P and slope = $-\frac{a}{b}$ is

⇒ y - y₁ = m(x - x₁)

⇒ y - (-b) = $-\frac{a}{b}$(x - 0)

⇒ b(y + b) = -ax

⇒ by + b² = -ax

⇒ ax + by + b² = 0

Hence, equation of required line is ax + by + b² = 0.

(i) By formula,

Slope = $\frac{y_2 - y_1}{x_2 - x_1}$

Substituting values we get,

$$\begin{aligned}\text{Slope of AB } = \frac{3 - 4}{3 - (-5)} \\[1em] = -\frac{1}{8}.\end{aligned}$$

By point-slope form,

Equation of AB is :

⇒ y - y₁ = m(x - x₁)

⇒ y - 4 = $-\frac{1}{8}[x - (-5)]$

⇒ 8(y - 4) = -1(x + 5)

⇒ 8y - 32 = -x - 5

⇒ x + 8y = -5 + 32

⇒ x + 8y = 27.

Hence, equation of AB is x + 8y = 27.

(ii) From part (i) above,

Slope of AB (m₁) = $-\frac{1}{8}$

Let slope of CD be m₂.

Since, AB ⊥ CD.

∴ m₁ × m₂ = -1

⇒ $-\frac{1}{8} \times m_2 = -1$

⇒ $m_2 = 8$.

From figure,

D = (-3, 0)

By point-slope form,

Equation of CD is :

⇒ y - y₁ = m(x - x₁)

⇒ y - 0 = 8[x - (-3)]

⇒ y = 8(x + 3)

⇒ y = 8x + 24.

Hence, equation of CD is y = 8x + 24.

Let co-ordinates of A be (a, 0) and B be (0, b).

Given,

BP = 2PA

$\frac{BP}{PA} = \frac{2}{1} \text{ or } \frac{PA}{BP} = \frac{1}{2}$.

∴ P divides AB in the ratio 1 : 2.

By section formula,

$$\begin{aligned}\Rightarrow P = \Big(\frac{m_1x_2 + m_2x_1}{m_1 + m_2}, \frac{m_1y_2 + m_2y_1}{m_1 + m_2}\Big) \\[1em] \Rightarrow (4, 3) = \Big(\frac{1 \times 0 + 2 \times a}{1 + 2}, \frac{1 \times b + 2 \times 0}{1 + 2}\Big) \\[1em] \Rightarrow (4, 3) = \Big(\frac{0 + 2a}{3}, \frac{b + 0}{3}\Big) \\[1em] \Rightarrow (4, 3) = \Big(\frac{2a}{3}, \frac{b}{3}\Big) \\[1em] \Rightarrow 4 = \frac{2a}{3} \text{ and } 3 = \frac{b}{3} \\[1em] \Rightarrow a = \frac{4 \times 3}{2} \text{ and } b = 9 \\[1em] \Rightarrow a = 6 \text{ and } b = 9.\end{aligned}$$

A = (a, 0) = (6, 0) and B = (0, b) = (0, 9).

Slope of AB = $\frac{9 - 0}{0 - 6} = -\frac{9}{6} = -\frac{3}{2}$.

By point-slope form,

Equation of AB is :

⇒ y - y₁ = m(x - x₁)

⇒ y - 0 = $-\frac{3}{2}$(x - 6)

⇒ 2y = -3(x - 6)

⇒ 2y = -3x + 18

⇒ 3x + 2y = 18.

Hence, equation of AB is 3x + 2y = 18.

In △XZB and △MZY,

∠XZB = ∠MZY (Vertically opposite angles are equal)

∠XBZ = ∠YMZ (Each = 60°)

△XZB ~ △MZY [By AA postulate]

In similar triangles,

The ratios between the lengths of corresponding sides are equal.

∴ $\frac{ZB}{MZ} = \frac{BX}{MY}$

As sides of equilateral triangle are equal so,

BX = AB

and

MY = MB

∴ $\frac{BX}{MY} = \frac{AB}{MB}$

∴ $\frac{ZB}{MZ} = \frac{BX}{MY} = \frac{AB}{MB}$ ......(1)

As, M is the mid-point of AB.

∴ MB = $\frac{1}{2}AB$

From (1),

$$\begin{aligned}\Rightarrow \frac{ZB}{MZ} = \frac{AB}{MB} \\[1em] \Rightarrow \frac{ZB}{AZ - AM} = \frac{AB}{\frac{1}{2}AB} \\[1em] \Rightarrow \frac{ZB}{AZ - AM} = 2 \\[1em] \Rightarrow ZB = 2(AZ - AM) \\[1em] \Rightarrow ZB = 2AZ - 2AM \\[1em] \Rightarrow ZB = 2AZ - AB \space [\because \text{M is mid-point of AB}] \\[1em] \Rightarrow ZB = AZ + AZ - AB \\[1em] \Rightarrow ZB = AZ - (AB - AZ) \\[1em] \Rightarrow ZB = AZ - ZB \\[1em] \Rightarrow 2ZB = AZ.\end{aligned}$$

Hence, proved that AZ = 2ZB.

In △ABQ and △ACK,

CK || BQ

∴ ∠ACK = ∠ABQ [Corresponding angles are equal]

∠KAC = ∠QAB (Common)

∴ △ACK ~ △ABQ

From figure,

AB = AC + CB = 3 + 6 = 9 cm.

In similar triangles,

The ratios between the lengths of corresponding sides are equal.

$$\begin{aligned}\therefore \frac{AC}{AB} = \frac{AK}{AQ} = \frac{CK}{BQ} \\[1em] \Rightarrow \frac{3}{9} = \frac{AK}{AQ} = \frac{CK}{7.5} \\[1em] \Rightarrow \frac{AK}{AQ} = \frac{3}{9} \\[1em] \Rightarrow \frac{AK}{AQ} = \frac{1}{3} \space ..........(1)\end{aligned}$$

Also,

$$\begin{aligned}\Rightarrow \frac{CK}{7.5} = \frac{1}{3} \\[1em] \Rightarrow CK = \frac{7.5}{3} = 2.5\end{aligned}$$

In △KRQ and △APQ,

AP || KR

∴ ∠KRQ = ∠APQ [Corresponding angles are equal]

∠KQR = ∠AQP (Common)

∴ △KRQ ~ △APQ

In similar triangles,

The ratios between the lengths of corresponding sides are equal.

$$\begin{aligned}\therefore \frac{KR}{AP} = \frac{KQ}{AQ} \\[1em] \Rightarrow \frac{KR}{4.5} = \frac{AQ - AK}{AQ} \\[1em] \Rightarrow \frac{KR}{4.5} = \frac{AQ}{AQ} - \frac{AK}{AQ} \\[1em] \Rightarrow \frac{KR}{4.5} = 1 - \frac{1}{3} \text{ (From 1)} \\[1em] \Rightarrow \frac{KR}{4.5} = \frac{2}{3} \\[1em] \Rightarrow KR = 4.5 \times \frac{2}{3} \\[1em] \Rightarrow KR = 3 \text{ cm}.\end{aligned}$$

From figure,

CR = CK + KR = 2.5 + 3 = 5.5 cm

Hence, CR = 5.5 cm

(i) Given,

⇒ BO : OD = 4 : 7

⇒ OD : OB = 7 : 4

Since,

△AOD and △AOB have common vertex at A and their bases OD and OB are along the same straight line.

$\therefore \frac{\text{Area of △AOD}}{\text{Area of △AOB}} = \frac{OD}{OB} = \frac{7}{4}$.

Hence, △AOD : △AOB = 7 : 4.

(ii) Since,

△AOB and △ACB have common vertex at A and their bases AO and AC are along the same straight line.

$\therefore \frac{\text{Area of △AOB}}{\text{Area of △ACB}} = \frac{AO}{AC} = \frac{4}{11}$

Hence, △AOB : △ACB = 4 : 11.

(iii) Given,

BO : OD = 4 : 7

In △AOB and △DOC

∠AOB = ∠COD [Vertically opposite angles are equal]

∠ABO = ∠ODC [Alternate angles are equal]

∴ △AOB ~ △DOC [By AA axiom]

By area theorem,

Ratio of area of two similar triangles is equal to the square of the ratio of the corresponding sides.

$$\begin{aligned}\therefore \frac{\text{Area of △DOC}}{\text{Area of △AOB}} = \Big(\frac{OD}{OB}\Big)^2 \\[1em] = \Big(\frac{7}{4}\Big)^2 \\[1em] = \frac{49}{16} \\[1em] = 49 : 16.\end{aligned}$$

Hence, △DOC : △AOB = 49 : 16.

(iv) Given,

BO : OD = 4 : 7

$$\begin{aligned}\Rightarrow \frac{BO}{OD} = \frac{4}{7} \\[1em] \Rightarrow \frac{OD}{BO} = \frac{7}{4} \\[1em] \Rightarrow \frac{OD}{BO} + 1 = \frac{7}{4} + 1 \\[1em] \Rightarrow \frac{OD + BO}{BO} = \frac{7 + 4}{4} \\[1em] \Rightarrow \frac{BD}{BO} = \frac{11}{4}.\end{aligned}$$

Since,

△ABD and △BOC have common vertex at B and their bases BD and OB are along the same straight line.

$\therefore \frac{\text{Area of △ABD}}{\text{Area of △BOC}} = \frac{BD}{BO} = \frac{11}{4}.$

Hence, △ABD : △BOC = 11 : 4.

Given,

Scale of model = 1 : 160

Scale factor (k) = $\frac{1}{160}$

(i) By formula,

⇒ Length of model = Scale factor × Length of ship

⇒ 1.2 = $\frac{1}{160}$ × Length of ship

⇒ Length of ship = 1.2 × 160 = 192 m.

Hence, length of ship = 192 m.

(ii) By formula,

⇒ Area of model = (Scale factor)² × Area of ship

⇒ 1.2 = $\Big(\frac{1}{160}\Big)^2$ × Area of ship

⇒ 1.2 = $\frac{1}{160 \times 160}$ × Area of ship

⇒ Area of ship = 1.2 × 160 × 160 = 30720 m².

Hence, area of ship = 30720 m².

(iii) By formula,

⇒ Volume of model = (Scale factor)³ × Volume of ship

⇒ 1.2 = $\Big(\frac{1}{160}\Big)^3$ × Volume of ship

⇒ 1.2 = $\frac{1}{160 \times 160 \times 160}$ × Volume of ship

⇒ Volume of ship = 1.2 × 160 × 160 × 160 = 49,15,200 m³.

Hence, volume of ship = 49,15,200 m³.

Given,

⇒ 4BE = 3EC

⇒ $\frac{BE}{EC} = \frac{3}{4}$

In △DFG and △DAB,

⇒ ∠DFG = ∠DAB [As corresponding angles are equal]

⇒ ∠FDG = ∠ADB (Common)

⇒ △DFG ~ △DAB [By AA postulate]

In similar triangles,

The ratios between the lengths of corresponding sides are equal.

$\frac{DF}{DA} = \frac{FG}{AB}$ ..........(1)

In trapezium ABCD, we have

EF || AB || DC

$$\begin{aligned}\therefore \frac{AF}{DF} = \frac{BE}{EC} \\[1em] \Rightarrow \frac{AF}{DF} = \frac{3}{4} \\[1em] \Rightarrow \frac{AF}{DF} + 1 = \frac{3}{4} + 1 \\[1em] \Rightarrow \frac{AF + DF}{DF} = \frac{3 + 4}{4} \\[1em] \Rightarrow \frac{AD}{DF} = \frac{7}{4} \\[1em] \Rightarrow \frac{DF}{AD} = \frac{4}{7} \space ..........(2)\end{aligned}$$

From (1) and (2), we get :

$\frac{FG}{AB} = \frac{4}{7}$

FG = $\frac{4}{7}$AB .........(3)

In △BEG and △BCD,

⇒ ∠BEG = ∠BCD [As corresponding angles are equal]

⇒ ∠GBE = ∠DBC (Common angle)

⇒ △BEG ~ △BCD [By AA postulate]

In similar triangles,

The ratios between the lengths of corresponding sides are equal.

$\frac{BE}{BC} = \frac{EG}{CD}$ ..............(4)

Given,

$$\begin{aligned}\Rightarrow \frac{BE}{EC} = \frac{3}{4} \\[1em] \Rightarrow \frac{EC}{BE} = \frac{4}{3} \\[1em] \Rightarrow \frac{EC}{BE} + 1 = \frac{4}{3} + 1 \\[1em] \Rightarrow \frac{EC + BE}{BE} = \frac{4 + 3}{3} \\[1em] \Rightarrow \frac{BC}{BE} = \frac{7}{3} \\[1em] \Rightarrow \frac{BE}{BC} = \frac{3}{7} \\[1em] \Rightarrow \frac{EG}{CD} = \frac{3}{7} \space [\text{From (4)}] \\[1em] \Rightarrow EG = \frac{3}{7}CD \\[1em] \Rightarrow EG = \frac{3}{7} \times 2AB \\[1em] \Rightarrow EG = \frac{6}{7}AB \space ..........(5)\end{aligned}$$

Adding equation (3) and (5),

$$\begin{aligned}\Rightarrow FG + EG = \frac{4}{7}AB + \frac{6}{7}AB \\[1em] \Rightarrow EF = \frac{10}{7}AB \\[1em] \Rightarrow 7EF = 10AB.\end{aligned}$$

Hence, proved that 7EF = 10AB.

Since, ABCD is a rhombus.

So, AD || BC.

In △DPA and △RPC,

⇒ ∠DPA = ∠RPC (Vertically opposite angles are equal)

⇒ ∠PAD = ∠PCR [Since, AD || CD and AC is transversal]

∴ △DPA ~ △RPC [By AA]

Since, corresponding sides of similar triangles are proportional we have :

⇒ $\frac{DP}{PR} = \frac{AD}{CR}$ ..........(1)

In rhombus all sides are equal.

∴ AD = DC

Substituting in (1) we get,

⇒ $\frac{DP}{PR} = \frac{DC}{CR}$

⇒ DP × CR = DC × PR.

Hence, proved that DP × CR = DC × PR.

Given, FB = FD

∴ ∠FBD = ∠FDB [Since, opposite sides of equal angles are equal] ......(1)

In △AED and △FCB,

∠ADE = ∠FBC [From 1]

∠AED = ∠FCB [Both = 90°]

∴ △AED ~ △FCB [By AA]

Since, corresponding sides of similar triangles are proportional we have :

⇒ $\frac{FB}{AD} = \frac{BC}{ED}$.

Hence, proved that $\frac{FB}{AD} = \frac{BC}{ED}$.

In △PQR and △PAB,

∠PQR = ∠PAB [Corresponding angles are equal]

∠PRQ = ∠PBA [Corresponding angles are equal]

Hence, △PQR ~ △PAB [By AA]

Since, corresponding sides of similar triangles are proportional to each other.

$\therefore \frac{PQ}{PA} = \frac{PR}{PB}$ ........(1)

In △PDR and △PQB,

∠PDR = ∠PQB [Corresponding angles are equal]

∠PRD = ∠PBQ [Corresponding angles are equal]

Hence, △PDR ~ △PQB [By AA]

Since, corresponding sides of similar triangles are proportional to each other.

$\therefore \frac{PD}{PQ} = \frac{PR}{PB}$ ........(2)

From (1) and (2) we get :

$$\begin{aligned}\Rightarrow \frac{PQ}{PA} =\frac{PD}{PQ} \\[1em] \Rightarrow PQ^2 = PD \times PA.\end{aligned}$$

Hence, proved that PQ² = PD × PA.

(i) Since, △APB and △CPB have common vertex at B and their bases AP and PC are along the same straight line AC.

∴ △APB : △CPB = AP : PC = 3 : 5.

Hence, △APB : △CPB = 3 : 5.

(ii) In △DPC and △APB,

∠DPC = ∠APB [Vertically opposite angles are equal]

∠PAB = ∠PCD [Alternate angles are equal]

∴ △DPC ~ △APB.

We know that,

The ratio of the areas of two similar triangles is equal to the ratio of squares of their corresponding sides.

$$\begin{aligned}\therefore \frac{\text{Area of △DPC}}{\text{Area of △APB}} = \frac{PC^2}{AP^2}\\[1em] = \frac{5^2}{3^2} = \frac{25}{9}.\end{aligned}$$

Hence, △DPC : △APB = 25 : 9.

(iii) Since, △DPC ~ △APB,

$\therefore \frac{PD}{PB} = \frac{PC}{AP} = \frac{5}{3}$ ..........(1)

Since, △ADP and △APB have common vertex at A and their bases DP and PB are along the same straight line.

∴ △ADP : △APB = DP : PB = 5 : 3 .........[From 1]

Hence, △ADP : △APB = 5 : 3.

(iv) From part (iii)

$\frac{PD}{PB} = \frac{5}{3}$

Let PD = 5x and PB = 3x.

From figure,

BD = PD + PB = 5x + 3x = 8x.

Since, △APB and △ADB have common vertex at A and their bases PB and DB are along the same straight line.

∴ △APB : △ADB = PB : BD = 3x : 8x = 3 : 8

Hence, △APB : △ADB = 3 : 8.

Join A and R. Let AR meet BQ at point D.

In ∆ACR, BD || CR.

By Basic proportionality theorem, we get

$\frac{AB}{BC} = \frac{AD}{DR}$ .....(1)

In ∆APR, DQ || AP.

By Basic proportionality theorem, we get

$\frac{PQ}{QR} = \frac{AD}{DR}$ .....(2)

From (1) and (2) we get :

$\frac{AB}{BC} = \frac{PQ}{QR}$

Hence proved that $\frac{AB}{BC} = \frac{PQ}{QR}$.

(i) In △ADC and △BEC,

∠ADC = ∠BEC [Both = 90°]

∠ACD = ∠BCE [Common angle]

∴ △ADC ~ △BEC [By AA].

Hence, proved that △ADC ~ △BEC.

(ii) Since, △ADC ~ △BEC.

We know that,

Corresponding sides of similar triangles are equal.

$\therefore \frac{CA}{CB} = \frac{CD}{CE}$

⇒ CA × CE = CB × CD.

Hence, proved that CA × CE = CB × CD.

(iii) From part (ii) we get,

$$\begin{aligned}\Rightarrow \frac{CA}{CB} = \frac{CD}{CE} \\[1em] \Rightarrow \frac{CA}{CD} = \frac{CB}{CE}\end{aligned}$$

∠DCE = ∠BCA [Common angle]

∴ △ABC ~ △DEC [By SAS]

Hence, proved that △ABC ~ △DEC.

(iv) Since, △ABC ~ △DEC

We know that,

Corresponding sides of similar triangles are proportional to each other.

⇒ $\frac{CA}{CD} = \frac{AB}{DE}$

⇒ CD × AB = CA × DE.

Hence, proved that CD × AB = CA × DE.

(i) Join AE.

In △EAD and △EBD,

⇒ ED = ED (Common)

⇒ AD = BD (As D is the mid-point of AB)

⇒ ∠EDA = ∠EDB (Both = 90°).

∴ △EAD ≅ △EBD (By SAS axiom)

∴ EA = EB (By C.P.C.T.)

Hence, proved that E is equidistant from A and B.

(ii) Draw FL ⊥ BC and FM ⊥ AB.

In △BLF and △BMF,

⇒ BF = BF (Common)

⇒ ∠LBF = ∠FBM (As BF is the angle bisector of ∠ABC)

⇒ ∠BLF = ∠BMF (Both = 90°).

∴ △BLF ≅ △BMF (By ASA axiom)

∴ FL = FM (By C.P.C.T.)

Hence, proved that F is equidistant from AB and BC.

Steps of construction :

1. Plot the points A(1, 1), B(5, 3) and C(2, 7).

2. Join the points AB, BC, and AC to form a triangle.

3. Draw DE, perpendicular bisector of AB. (As locus of points equidistant from two points is the perpendicular bisector of line joining them).

4. Draw AF, angle bisector of A. (As locus of points equidistant from two lines is the angular bisector of angle between them).

5. Mark point P as the intersection of DE and AF.

6. Measure AP.

Hence, AP = 2.8 cm.

Since, speed of sound is 332 m/s.

So, in 1 s it will travel = 332 meters.

Hence, locus of all the people on Earth's surface who will hear sound one second later will be the circumference of a circle whose radius is 332 m and centre is the point where the gun is fired.

Since, ABCD is a cyclic quadrilateral and sum of opposite angles in a cyclic quadrilateral = 180°.

∴ ∠ADC + ∠ABC = 180°

⇒ 130° + ∠ABC = 180°

⇒ ∠ABC = 50°.

From figure,

⇒ ∠FBC = ∠ABC = 50°.

In △FBC,

By angle sum property of triangle,

⇒ ∠FBC + ∠BCF + ∠CFB = 180°

⇒ 50° + ∠BCF + 90° = 180°

⇒ ∠BCF = 180° - 90° - 50° = 40°.

Given,

BC = BE

In △BCE,

⇒ ∠BEC = ∠BCE = 40°. (As angles opposite to equal sides are equal)

In △FBE,

By angle sum property of triangle,

⇒ ∠BEF + ∠FBE + ∠EFB = 180°

⇒ 40° + ∠FBE + 90° = 180°

⇒ ∠FBE = 180° - 90° - 40° = 50°.

From figure,

∠CBE = ∠FBC + ∠FBE = 50° + 50° = 100°.

Hence, ∠CBE = 100°.

In △AOB,

⇒ OA = OB [Radius of same circle]

⇒ ∠OBA = ∠BAO = 30° [Angles opposite to equal sides are equal]

Also,

⇒ ∠OBA + ∠BAO + ∠AOB = 180° [By angles sum property of triangle]

⇒ 30° + 30° + ∠AOB = 180°

⇒ ∠AOB = 180° - 60° = 120°

In △OCB,

OC = OB [Radius of same circle]

⇒ ∠OBC = ∠OCB = 57° [Angles opposite to equal sides are equal]

Also,

⇒ ∠OCB + ∠OBC + ∠BOC = 180° [By angles sum property of triangle]

⇒ 57° + 57° + ∠BOC = 180°

⇒ ∠BOC = 180° - 114° = 66°

From figure,

⇒ ∠AOB = ∠AOC + ∠BOC

⇒ 120° = ∠AOC + 66°

⇒ ∠AOC = 120° - 66° = 54°.

Hence, ∠AOC = 54° and ∠BOC = 66°.

Let AB = AC = x

Given,

OP ⊥ AB and OQ ⊥ AC

⇒ OM ⊥ AB and ON ⊥ AC

Since, the perpendicular to a chord from the center of the circle bisects the chord,

∴ AM = MB = $\frac{x}{2}$ and AN = NC = $\frac{x}{2}$

∴ MB = NC .............(1)

Since, equal chords of a circle are equidistant from the center,

∴ ON = OM = y (let)

Let radius of circle be r.

From figure,

⇒ OQ = OP = r

⇒ QN = OQ - ON = r - y

⇒ PM = OP - OM = r - y

∴ QN = PM .........(2)

In △QNC and △PMB,

⇒ NC = MB [From (1)]

⇒ QN = PM [From (2)]

⇒ ∠QNC = ∠PMB [Both equal to 90°]

∴ △QNC ≅ △∠PMB by SAS axiom.

∴ PB = QC (By C.P.C.T.)

Hence, proved that PB = QC.

(i) Arc AX subtends ∠AOX at the center and ∠APX at the remaining part of the circle.

⇒ ∠AOX = 2∠APX (∵ angle subtended on center is twice the angle subtended on the remaining part of the circle.)

⇒ ∠AOX = 2 × 30° = 60°.

Hence, the value of ∠AOX = 60°.

(ii) From figure,

∠XPY = 90° [∵ angle in semicircle = 90°]

∴ ∠APY = ∠XPY - ∠APX = 90° - 30° = 60°.

Hence, the value of ∠APY = 60°.

(iii) From figure,

∠APB = 90° [∵ angle in semicircle = 90°]

∴ ∠BPY = ∠APB - ∠APY = 90° - 60° = 30°.

Hence, the value of ∠BPY = 30°.

(iv) In △AOX,

OA = OX (Radius of same circle)

∴ ∠OAX = ∠OXA

Since, sum of angles of a triangle = 180°

∴ ∠AOX + ∠OAX + ∠OXA = 180°

⇒ 60° + ∠OAX + ∠OAX = 180°

⇒ 2∠OAX = 120°

⇒ ∠OAX = $\frac{120°}{2}$

⇒ ∠OAX = 60°.

Hence, the value of ∠OAX = 60°.

In △ABD,

AB = AD

∠ADB = ∠ABD (∵ angles opposite to equal sides are equal) ......(1)

In △BDC,

BD = CD

∠DCB = ∠DBC (∵ angles opposite to equal sides are equal) ......(2)

In △ADB,

⇒ ∠DAB + ∠ADB + ∠ABD = 180° [Angle sum property of triangle]

⇒ ∠DAB + ∠ABD + ∠ABD = 180° [From (1)]

⇒ ∠DAB + 2∠ABD = 180°

⇒ ∠DAB + ∠DBC = 180° [As, ∠DBC = 2∠ABD (Given)]

⇒ ∠DAB + ∠DCB = 180° [From (2)]

Since, ∠DAB and ∠DCB are opposite angles of a quadrilateral and sum of opposite angles in a cyclic quadrilateral = 180°.

Hence, proved that ABCD is a cyclic quadrilateral.

From figure,

∠ADB = 90° [Angle subtended by diameter is a right angle.]

In △COD,

OC = OD [Radius of same circle]

∴ OC = OD = CD.

So, △COD is an equilateral triangle.

∴ ∠COD = 60°.

We know that,

Angle subtended by chord in the center is double the angle in the circumference.

⇒ ∠COD = 2∠CAD

⇒ ∠CAD = $\frac{1}{2}$∠COD

⇒ ∠CAD = $\frac{1}{2} \times 60°$ = 30°.

In △APD,

⇒ ∠PAD = ∠CAD = 30°.

⇒ ∠ADP + ∠ADB = 180° [Linear pair]

⇒ ∠ADP = 180° - ∠ADB = 180° - 90° = 90°.

By angle sum property.

⇒ ∠ADP + ∠APD + ∠PAD = 180°

⇒ 90° + ∠APD + 30° = 180°

⇒ ∠APD + 120° = 180°

⇒ ∠APD = 180° - 120°

⇒ ∠APD = 60°.

From figure,

⇒ ∠APB = ∠APD = 60°.

Hence, proved that ∠APB = 60°.

(i) Since,

⇒ AB = AC

⇒ ∠ACB = ∠ABC = 72° [∵ angles opposite to equal sides are equal]

From figure,

⇒ ∠BAO = ∠ACB = 72° [Angles in alternate segment are equal]

⇒ OA = OB [Tangents from a fixed point outside the circle are equal.]

⇒ ∠OBA = ∠BAO = 72° [Since angle opposite to equal sides are equal]

In △ABO,

⇒ ∠BAO + ∠OBA + ∠AOB = 180° [By angle sum property of triangle]

⇒ 72° + 72° + ∠AOB = 180°

⇒ ∠AOB = 180° - 144° = 36°.

Hence, ∠AOB = 36°.

(ii) We know that,

Angle subtended by a chord at the centre of the circle is twice the angle subtended by it at any point of the circumference.

∴ ∠ADB = 2∠ACB = 2 × 72° = 144°.

Hence, angle subtended by AB at the centre of the circle = 144°.

From figure,

PQ = PR [Tangents from a fixed point outside the circle are equal.]

⇒ ∠PRQ = ∠PQR = a [Since angle opposite to equal sides are equal]

In △PQR,

⇒ ∠QPR + ∠PQR + ∠PRQ = 180° [By angle sum property of triangle]

⇒ 40° + a + a = 180°

⇒ 2a = 180° - 40°

⇒ 2a = 140°

⇒ a = $\frac{140°}{2}$

⇒ a = 70°.

From figure,

⇒ ∠PRQ + ∠QRT + ∠TRS = 180° [Linear pair]

⇒ 70° + 75° + ∠TRS = 180°

⇒ ∠TRS = 180° - 145° = 35°

⇒ SR = ST [Tangents from a fixed point outside the circle are equal.]

⇒ ∠STR = ∠TRS = 35° [Since angle opposite to equal sides are equal]

∴ c = 35°

In △SRT,

⇒ ∠STR + ∠TRS + ∠TSR = 180° [By angle sum property of triangle]

⇒ 35° + 35° + b = 180°

⇒ b = 180° - 70° = 110°.

Hence, a = 70°, b = 110° and c = 35°.

(i) As, BC is the diameter.

We know that,

Angle in a semicircle is a right angle.

∴ ∠BDC = 90°

From figure,

⇒ ∠BDC + ∠BDA = 180° [Linear Pair]

⇒ 90° + ∠BDA = 180°

⇒ ∠BDA = 180° - 90°

⇒ ∠BDA = 90°.

As, AB is the tangent and BC is diameter and tangent at any point and line from that point to center are perpendicular to each other.

⇒ ∠ABC = 90°

In △ABC and △ABD,

∠ABC = ∠ADB (Both equal to 90°)

∠BAD = ∠BAC (Common)

∴ △ABC ~ △ABD

In similar triangles,

Ratio of corresponding sides are in equal proportion.

$$\begin{aligned}\therefore \frac{AC}{AB} = \frac{AB}{AD} \\[1em] \Rightarrow AB^2 = AC \times AD.\end{aligned}$$

Hence, proved that AB² = AC × AD.

(ii) In △ABC and △BDC,

⇒ ∠ABC = ∠BDC (Both equal to 90°)

⇒ ∠BCA = ∠BCD (Common)

∴ △ABC ~ △BDC

In similar triangles,

Ratio of corresponding sides are in equal proportion.

$$\begin{aligned}\therefore \frac{AD}{BC} = \frac{BC}{CD} \\[1em] \Rightarrow BC^2 = AD \times CD.\end{aligned}$$

Hence, proved that BC² = AD × CD.

As, tangents from an exterior fixed point to a circle are equal in length,

Let,

AD = AF = x cm

BD = BE = y cm

CF = CE = z cm

From figure,

⇒ AD + BD = AB

⇒ x + y = 12 .........(1)

⇒ AF + CF = AC

⇒ x + z = 10 ........(2)

⇒ BE + CE = BC

⇒ y + z = 8 ..........(3)

Subtracting eq (3) from (1), we get :

⇒ x + y - (y + z) = 12 - 8

⇒ x - z = 4 ............(4)

Adding equation (2) and (4), we get :

⇒ (x + z) + (x - z) = 10 + 4

⇒ 2x = 14

⇒ x = $\frac{14}{2}$

⇒ x = 7.

Substituting value of x in equation (1), we get :

⇒ 7 + y = 12

⇒ y = 12 - 7

⇒ y = 5.

Substituting value of y in equation (3), we get :

⇒ y + z = 8

⇒ 5 + z = 8

⇒ z = 8 - 5

⇒ z = 3.

∴ AD = x = 7, BE = y = 5 and CF = z = 3.

Hence, AD = 7 cm, BE = 5 cm and CF = 3 cm.

(i) From figure,

⇒ AB = AP + PB

⇒ 24 = 4 + PB

⇒ PB = 24 - 4

⇒ PB = 20 cm.

We know that,

If two chords intersect in a circle, then the products of the measures of the segments of the chords are equal.

⇒ AP × PB = CP × PD

⇒ 4 × 20 = CP × 8

⇒ 80 = 8CP

⇒ CP = $\frac{80}{8}$ = 10.

Hence, CP = 10 cm.

(ii) Let CP = x, so PD = (6.5 - x).

We know that,

If two chords intersect in a circle , then the products of the measures of the segments of the chords are equal.

⇒ AP × PB = CP × PD

⇒ 3 × 2.5 = x(6.5 - x)

⇒ 7.5 = 6.5x - x²

⇒ x² - 6.5x + 7.5 = 0

⇒ x² - 5x - 1.5x + 7.5 = 0

⇒ x(x - 5) - 1.5(x - 5) = 0

⇒ (x - 1.5)(x - 5) = 0

⇒ x - 1.5 = 0 or x - 5 = 0

⇒ x = 1.5 or x = 5.

Hence, CP = 1.5 cm or 5 cm.

(i) We know that,

If two chords of a circle intersect externally, then the product of the lengths of the segments are equal.

⇒ AP × PB = CP × PD

⇒ 8 × PB = 5 × 4

⇒ PB = $\frac{20}{8}$ = 2.5

From figure,

AB = PA - PB = 8 - 2.5 = 5.5 cm.

Hence, AB = 5.5 cm.

(ii) From figure,

⇒ PC = CD + PD

⇒ 30 = 14 + PD

⇒ PD = 16 cm.

We know that,

If two chords of a circle intersect externally, then the product of the lengths of the segments are equal.

⇒ AP × PB = PC × PD

⇒ 24 × PB = 30 × 16

⇒ PB = $\frac{30 \times 16}{24}$ = 20

From figure,

AB = PA - PB = 24 - 20 = 4 cm.

Hence, AB = 4 cm.

Draw tangent at point P,

∠PAB = ∠BPS .............(1) [Angles in alternate segments are equal.]

∠PCD = ∠DPS .............(2) [Angles in alternate segments are equal.]

Subtracting (1) from (2), we get :

⇒ ∠PCD - ∠PAB = ∠DPS - ∠BPS ............(3)

In △PAC,

⇒ ∠PCD = ∠PAC + ∠CPA [An exterior angle is equal to sum of two opposite interior angles.]

⇒ ∠PCD = ∠PAB + ∠CPA [From figure, ∠PAC = ∠PAB]

Substituting above value of ∠PCD in (3), we get :

⇒ ∠PAB + ∠CPA - ∠PAB = ∠DPS - ∠BPS

⇒ ∠CPA = ∠DPB [From figure, ∠DPS - ∠BPS = ∠DPB].

Hence, proved that ∠CPA = ∠DPB.

We know that,

If a chord and a tangent intersect externally, then the product of the lengths of the segments of the chord is equal to the square of the length of the tangent from the point of contact to the point of intersection.

∴ AP² = PM x PN ..........(1)

From P, PB is the tangent and PMN is the secant for second circle.

∴ PB² = PM x PN ......... (2)

From (1) and (2), we have

⇒ AP² = PB²

⇒ AP = PB

Hence, proved that P is the midpoint of AB.

(i) Since, PAT || BC

∠BAP = ∠ABC [Alternate angles are equal] .........(1)

In cyclic quadrilateral ABCD,

∠ABC + ∠ADC = 180° [Sum of opposite angles in a cyclic quadrilateral = 180°] ................(2)

From figure,

∠ADQ + ∠ADC = 180° [Linear pairs] .........(3)

From (2) and (3), we get :

∠ADQ = ∠ABC ............(4)

From (1) and (4), we get :

∠BAP = ∠ADQ.

Hence, proved that ∠BAP = ∠ADQ.

(ii) We know that,

Angle subtended by an arc at the center is twice the angle subtended at any other point of circumference.

Arc AB subtends ∠AOB at the center and ∠ADB at the remaining part of the circle.

⇒ ∠AOB = 2∠ADB ............(3)

From figure,

∠ADB = ∠PAB [Angles in alternate segment are equal.]

Substituting above value in (3), we get :

⇒ ∠AOB = 2∠PAB

∠PAB = ∠ADQ [Proved above]

⇒ ∠AOB = 2∠ADQ.

Hence, proved that ∠AOB = 2∠ADQ.

(iii) From figure,

∠BAP = ∠ADB [Angles in alternate segment are equal.] .........(4)

∠BAP = ∠ADQ [Proved in part (i)] ...........(5)

From (4) and (5), we get :

∠ADQ = ∠ADB.

Hence, proved that ∠ADQ = ∠ADB.

Steps of construction :

1. Draw a line segment AC = 5 cm.

2. From A draw an arc of radius 6 cm and from C draw an arc of radius 7 cm, intersecting each other at point B.

3. Draw XY, the perpendicular bisector of AB, and mark D as the mid point.

4. Draw perpendicular bisector of CD.

5. From C, draw a line segment perpendicular to AC, which intersect the PQ, perpendicular bisector of CD at O.

6. Take O as centre and OD as radius draw a circle.

Steps of construction :

1. Taking O as center and radius = 4 cm draw a circle.

2. Draw AB, the diameter of circle with center O.

3. Produce AB to X such that BX = 4 cm.

4. Draw DE perpendicular bisector of BX, intersecting BX at C.

5. From C, as center and CX as radius draw a circle touching circle with center O at B and passing through X.

(i) If distance covered in one revolution is 8 cm, then distance covered in 9 revolutions = 9 x 8 = 72 cm or distance covered in 1 second = 72 cm.

Distance covered in $4\frac{1}{2}$ seconds = 72 x 4.5 = 324 cm.

Hence, distance travelled by the cylinder in $4\frac{1}{2}$ seconds is 324 cm.

Figure unavailable in the source: A cylinder of circumference 8 cm and length 21 cm rolls without sliding for seconds at the rate of 9 complete rounds per second. Find: (i) distance travelled by the cylinder in seconds, and (ii) the area covered by the cylinder in seconds…

(ii) Given, circumference = 8 cm.

∴ 2πr = 8

⇒ $2 \times \frac{22}{7} \times r = 8$

⇒ r = $\frac{8 \times 7}{2 \times 22} = \frac{14}{11}$ cm.

By formula,

Curved surface area = 2πrh

= 2 x $\frac{22}{7} \times \frac{14}{11}$ x 21

= 2 x 2 x 2 x 21

= 168 cm².

So, the area covered in one revolution = 168 cm²,

∴ The area covered in 9 revolutions = 168 x 9 = 1512 cm²,

∴ The area covered in 1 second = 1512 cm²,

Thus, the area covered in $4\frac{1}{2}$ seconds = 1512 x 4.5 = 6804 cm².

Hence, the area covered by cylinder in $4\frac{1}{2}$ seconds = 6804 cm².

Given,

Diameter of cylindrical bucket = 28 cm

Radius (r) = $\frac{\text{Diameter}}{2} = \frac{28}{2}$ = 14 cm.

Height (H) = 72 cm.

Length of rectangular tank (l) = 66 cm

Breadth of rectangular tank (b) = 28 cm.

Let height of water be h cm in the rectangular tank.

Since, water is emptied into a rectangular tank from cylindrical bucket,

∴ Volume of cylindrical bucket = Volume of water in rectangular tank

$$\begin{aligned}\Rightarrow πr^2H = l × b × h \\[1em] \Rightarrow h = \frac{πr^2H}{lb} \\[1em] \Rightarrow h = \frac{\frac{22}{7} \times 14^2 \times 72}{66 \times 28} \\[1em] \Rightarrow h = \frac{22 \times 196 \times 72}{7 \times 66 \times 28} \\[1em] \Rightarrow h = \frac{7 \times 72}{7 \times 3} \\[1em] \Rightarrow h = 24\text{ cm}.\end{aligned}$$

Hence, height of water level in rectangular tank = 24 cm.

Let the radius of a solid right cylinder be r cm and height be h cm.

∴ Volume of cylinder = πr²h

New radius (r') = r + $\frac{20}{100} \times r = \frac{120r}{100}$ = 1.2r.

New height (h') = h - $\frac{20}{100} \times h = \frac{80h}{100}$ = 0.8h.

⇒ New volume of the solid right circular cylinder = πr'²h'

= π x (1.2r)² x 0.8h

= 1.152 πr²h.

Increase in volume = New volume - Original volume

= 1.152 πr²h - πr²h

= 0.152 πr²h.

By formula,

Percentage change in volume = $\frac{\text{Increase in volume}}{\text{Original volume}}$ x 100%

= $\frac{0.152πr^2h}{πr^2h}$ x 100%

= 15.2 %.

Hence, percentage change in volume = 15.2 %.

Given,

Sum of the inner and the outer curved surfaces of a hollow metallic cylinder = 1056 cm²

∴ 2πrh + 2πRh = 1056

⇒ 2πh(r + R) = 1056

⇒ $\frac{44}{7}h(r + R)$ = 1056

⇒ h(r + R) = $\frac{1056 \times 7}{44}$

⇒ h(r + R) = 168 ...........(1)

Volume of the material = 1056 cm³

∴ π(R² - r²)h = 1056

⇒ $\frac{22}{7} \times (R^2 - r^2)h$ = 1056

⇒ (R² - r²)h = $\frac{1056 \times 7}{22}$

⇒ (R² - r²)h = 336 .........(2)

Dividing (2) by (1) we get,

$$\begin{aligned}\Rightarrow \frac{(R^2 - r^2)h}{(R + r)h} = \frac{336}{168} \\[1em] \Rightarrow \frac{(R - r)(R + r)h}{(R + r)h} = 2 \\[1em] \Rightarrow R - r = 2 \\[1em] \Rightarrow R = r + 2\end{aligned}$$

Substituting value of height and R in equation (1) we get :

⇒ 21(r + r + 2) = 168

⇒ 21(2r + 2) = 168

⇒ 2r + 2 = $\frac{168}{21}$

⇒ 2r + 2 = 8

⇒ 2r = 8 - 2

⇒ 2r = 6

⇒ r = $\frac{6}{2}$

⇒ r = 3 cm.

⇒ R = r + 2 = 3 + 2 = 5 cm.

Hence, internal radius = 3 cm and external radius = 5 cm.

Given,

Height of cylindrical portion (h) = 3 m

Height of conical part (H) = Maximum height of tent (above ground) - Height of cylindrical portion = 13.5 - 3 = 10.5 m.

Radius of base of cylindrical portion = Radius of conical portion = r = 14 m.

⇒ Curved surface area of tent = Curved surface area of cone + Curved surface area of cylinder

⇒ Curved surface area of tent (C) = πrl + 2πrh .......(1)

We know that,

$$\begin{aligned}l = \sqrt{r^2 + H^2} \\[1em] = \sqrt{14^2 + (10.5)^2} \\[1em] = \sqrt{196 + 110.25} \\[1em] = \sqrt{306.25} \\[1em] = 17.5 \text{ m}.\end{aligned}$$

Substituting values in equation (1), we get :

$$\begin{aligned}\Rightarrow C = \frac{22}{7} \times 14 \times 17.5 + 2 \times \frac{22}{7} \times 14 \times 3 \\[1em] = 22 \times 2 \times 17.5 + 2 \times 22 \times 2 \times 3 \\[1em] = 770 + 264 \\[1em] = 1034 \text{ m}^2.\end{aligned}$$

Given,

Cost of painting the inner surface of the tent at ₹ 4 per sq. meter.

Total cost = Curved surface area of tent × 4

= 1034 × 4 = ₹ 4136.

Hence, cost of painting the inner surface = ₹ 4136.

Given,

Diameter of copper wire = 6 mm = 0.6 cm

Length of cylinder (l) = 18 cm

Diameter of cylindrical base = 49 cm

Radius of cylindrical base (r) = $\frac{49}{2}$ = 24.5 cm.

(i) Given,

Copper wire is evenly wrapped on the cylinder to cover the whole surface.

As it is covers the whole length of cylinder.

∴ No. of times it wraps around = $\frac{\text{Length of cylinder}}{\text{Diameter of wire}} = \frac{18}{0.6}$ = 30.

For each turn, length of copper wire = Circumference of cylinder = 2πr

= 2 × $\frac{22}{7}$ × 24.5

= 154 cm.

Total length = 30 × 154 = 4620 cm = $\frac{4620}{100}$ = 46.20 m.

Hence, total length of copper wire = 46.20 m.

(ii) Since, wire is in the form of cylinder.

Height = Length of wire

Given,

Radius of wire (R) = $\frac{0.6}{2}$ = 0.3 cm.

Volume of wire = πR² × length

= $\frac{22}{7} \times (0.3)^2 \times 4620$

= $\frac{9147.6}{7}$ = 1306.8 cm³.

Hence, volume of wire = 1306.8 cm³.

Given,

Radius of pool (R) = 5 m,

Depth of pool (H) = 1.4 m.

Since, pool is in the form of cylinder.

Volume of water in pool = πR²H

= $\frac{22}{7} \times 5^2 \times 1.4$ = 110 m³

= 110 × 1000 litres = 11 × 10⁴ liters.

Volume of water coming out of filling pipe = 20 litres/sec.

Time taken to fill the tank = $\frac{11 \times 10^4}{20}$ second

= $\frac{11 \times 10^4}{20 \times 60}$ minutes

= $\frac{1100}{12} = 91\frac{2}{3}$ minutes.

Let radius of pipe be r cm.

Volume of water coming out from draining pipe in 1 sec = Area of cross section × Rate

= πr² × 2 m/sec

= (200 × πr²) cm³

= (200 × πr² × 10^-3) liters.

Given,

Time required for emptying the pool = 42 min = 42 × 60 seconds.

$$\begin{aligned}\therefore \frac{\text{Volume of water in pool}}{\text{Volume of water drained out in 1 sec}} = 42 \times 60 \\[1em] \Rightarrow \frac{11 \times 10^4}{200 \times 10^{-3} \times πr^2} = 42 \times 60 \\[1em] \Rightarrow \frac{11 \times 10^2 \times 10^3}{2 \times \frac{22}{7} \times r^2} = 42 \times 60 \\[1em] \Rightarrow \frac{1100 \times 7 \times 10^3}{44r^2} = 2520 \\[1em] \Rightarrow \frac{100 \times 7 \times 10^3}{4r^2} = 2520 \\[1em] \Rightarrow \frac{700000}{4r^2} = 2520 \\[1em] \Rightarrow r^2 = \frac{700000}{4 \times 2520} \\[1em] \Rightarrow r^2 = \frac{100000}{4 \times 360} \\[1em] \Rightarrow r^2 = \frac{2500}{36} \\[1em] \Rightarrow r = \sqrt{\frac{2500}{36}} \\[1em] \Rightarrow r = \frac{50}{6} = \frac{25}{3} = 8\frac{1}{3} \text{ cm}.\end{aligned}$$

Hence, time taken to fill the pool = $91\frac{2}{3}$ minutes and radius of pipe = $8\frac{1}{3}$ cm.

Let length of cylindrical part be h cm and radius of hemispherical and conical part be r cm.

Given,

Volume of water required to fill whole test tube = $\frac{2849}{3}$ cm³.

Volume of water required to fill test tube upto a level which is 2 cm below the top of tube = $\frac{2618}{3}$ cm³.

∴ Volume of water above 2 cm part of the test tube = $\frac{2849}{3} - \frac{2618}{3} = \frac{231}{3}$ = 77 cm³.

$$\begin{aligned}\Rightarrow π \times r^2 \times 2 = 77 \\[1em] \Rightarrow \frac{44}{7}r^2 = 77 \\[1em] \Rightarrow r^2 = \frac{77 \times 7}{44} \\[1em] \Rightarrow r^2 = \frac{539}{44} \\[1em] \Rightarrow r^2 = \frac{49}{4} \\[1em] \Rightarrow r = \sqrt{\frac{49}{4}} \\[1em] \Rightarrow r = \frac{7}{2} = 3.5 \text{ cm}.\end{aligned}$$

Volume of water in test tube = Volume of cylindrical part + Volume of hemispherical part = $\frac{2849}{3}$ cm³.

$$\begin{aligned}\therefore πr^2h + \frac{2}{3}πr^3 = \frac{2849}{3} \\[1em] \Rightarrow πr^2\Big(h + \frac{2r}{3}\Big) = \frac{2849}{3} \\[1em] \Rightarrow \frac{22}{7}r^2\Big(h + \frac{2r}{3}\Big) = \frac{2849}{3} \\[1em] \Rightarrow \frac{22}{7} \times (3.5)^2 \times \Big(h + \frac{2 \times 3.5}{3}\Big) = \frac{2849}{3} \\[1em] \Rightarrow 38.5 \times \Big(h + \frac{7}{3}\Big) = \frac{2849}{3} \\[1em] \Rightarrow h + \frac{7}{3} = \frac{2849}{3 \times 38.5} \\[1em] \Rightarrow h + \frac{7}{3} = \frac{2849}{115.5} \\[1em] \Rightarrow h + \frac{7}{3} = \frac{28490}{1155} \\[1em] \Rightarrow h + \frac{7}{3} = \frac{74}{3} \\[1em] \Rightarrow h = \frac{74}{3} - \frac{7}{3} \\[1em] \Rightarrow h = \frac{67}{3} = 22\frac{1}{3} \text{ cm}.\end{aligned}$$

Hence, radius of tube = 3.5 cm and length of cylindrical part = $22\frac{1}{3}$ cm.

Given,

Radius of cone (R) = 5 cm.

Height of cone (H) = 12 cm

Let radius of sphere be r cm.

From figure,

FE = FD = r cm

AD = 12 cm, DC = 5 cm

Slant height (AC) = $\sqrt{AD^2 + CD^2} = \sqrt{12^2 + 5^2}$

= $\sqrt{144 + 25} = \sqrt{169}$ = 13 cm.

In △ADC and △AFE,

⇒ ∠ADC = ∠AEF (Both equal to 90°)

⇒ ∠A = ∠A (Common)

∴ △ADC ~ △AFE (By A.A. axiom)

Ratio of corresponding sides of similar triangle are equal.

$$\begin{aligned}\therefore \frac{FE}{DC} = \frac{AF}{AC} \\[1em] \Rightarrow \frac{r}{5} = \frac{12 - r}{13} \\[1em] \Rightarrow 13r = 5(12 - r) \\[1em] \Rightarrow 13r = 60 - 5r \\[1em] \Rightarrow 18r = 60 \\[1em] \Rightarrow r = \frac{60}{18} = \frac{10}{3} \text{ cm}.\end{aligned}$$

Ratio of volume of sphere to volume of cone.

$$\begin{aligned}\Rightarrow \frac{\text{Volume of sphere}}{\text{Volume of cone}} = \frac{\frac{4}{3}πr^3}{\frac{1}{3}πR^2H} \\[1em] = \frac{4r^3}{R^2H} \\[1em] = \frac{4 \times \Big(\frac{10}{3}\Big)^3}{5^2 \times 12} \\[1em] = \frac{4 \times \frac{1000}{27}}{25 \times 12} \\[1em] = \frac{4000}{25 \times 27 \times 12} \\[1em] = \frac{40}{81} = 40 : 81.\end{aligned}$$

Hence, proved that ratio of volume of sphere and conical vessel = 40 : 81.

Given,

Height of cylinder (H) = 14 cm

Let outer radius of cylinder be R cm and inner radius be r cm.

Given,

Difference between the outer and the inner curved surface areas of a hollow cylinder is 88 sq. cm.

$$\begin{aligned}\therefore 2πRH - 2πrH = 88 \\[1em] \Rightarrow 2πH(R - r) = 88 \\[1em] \Rightarrow 2 \times \frac{22}{7} \times 14 (R - r) = 88 \\[1em] \Rightarrow 88(R - r) = 88 \\[1em] \Rightarrow R - r = 1 \space ............(1)\end{aligned}$$

Given,

Volume of metal = 176 cu. cm

$$\begin{aligned}\therefore π(R^2 - r^2)H = 176 \\[1em] \Rightarrow \frac{22}{7} \times (R^2 - r^2) \times 14 = 176 \\[1em] \Rightarrow 44(R^2 - r^2) = 176 \\[1em] \Rightarrow (R^2 - r^2) = \frac{176}{44} \\[1em] \Rightarrow (R^2 - r^2) = 4 \\[1em] \Rightarrow (R - r)(R + r) = 4 \\[1em] \Rightarrow 1 \times (R + r) = 4 \text{ .......[From (1)]} \\[1em] \Rightarrow (R + r) = 4 ...........(2)\end{aligned}$$

Adding equation (1) and (2), we get :

⇒ R - r + R + r = 1 + 4

⇒ 2R = 5

⇒ R = $\frac{5}{2}$ = 2.5 cm

Substituting value of R in equation (1), we get :

⇒ 2.5 - r = 1

⇒ r = 2.5 - 1 = 1.5 cm

Hence, outer radii = 2.5 cm and inner radii = 1.5 cm.

Let original radius be r units and new radius be R units.

Given,

Surface area of a solid sphere is increased by 21% without changing its shape.

$$\begin{aligned}\therefore 4πR^2 = 4πr^2 + \frac{21}{100} \times 4πr^2 \\[1em] \Rightarrow 4πR^2 = 4π(r^2 + \frac{21}{100}r^2) \\[1em] \Rightarrow R^2 = r^2 + \frac{21}{100}r^2 \\[1em] \Rightarrow R^2 = \frac{121}{100}r^2 \\[1em] \Rightarrow R = \sqrt{\frac{121}{100}r^2} \\[1em] \Rightarrow R = \frac{11}{10}r.\end{aligned}$$

By formula,

$$\begin{aligned}\text{Percentage increase in radius} = \frac{R - r}{r} \times 100 \\[1em] = \frac{\frac{11}{10}r - r}{r} \times 100 \\[1em] = \frac{\frac{r}{10}}{r} \times 100 \\[1em] = \frac{1}{10} \times 100 \\[1em] = 10%.\end{aligned}$$

Hence, percentage increase in radius = 10%.

(ii) Let original volume be v and new volume be V.

By formula,

$$\begin{aligned}\text{Percentage increase in volume} = \frac{V - v}{v} \times 100 \\[1em] = \frac{\frac{4}{3}πR^3 - \frac{4}{3}πr^3}{\frac{4}{3}πr^3} \times 100\\[1em] = \frac{\frac{4}{3}π(R^3 - r^3)}{\frac{4}{3}πr^3} \times 100\\[1em] = \frac{R^3 - r^3}{r^3} \times 100 \\[1em] = \frac{\Big(\frac{11}{10}r\Big)^3 - r^3}{r^3} \times 100 \\[1em] = \frac{\frac{1331}{1000} - 1}{1} \times 100 \\[1em] = \frac{\frac{1331 - 1000}{1000}}{1} \times 100 \\[1em] = \frac{331}{1000}\times 100 \\[1em] = 33.1%.\end{aligned}$$

Hence, percentage increase in volume = 33.1%.

Given,

Height of the cone (h) = 24 cm

Height of the cylinder (H) = 60 - 24 = 36 cm

Radius of the cone (r) = 10 cm

Given,

Radius of the base of the cone = Twice the radius of the base of the cylinder.

Radius of base of cylinder (R) = 5 cm.

By formula,

⇒ l² = r² + h²

⇒ l² = 10² + 24²

⇒ l² = 100 + 576

⇒ l² = 676

⇒ l = $\sqrt{676}$ = 26 cm.

Total surface area of the toy = Surface area of the conical part + Base area of conical part + Surface area of the cylinder

= πrl + πr² + 2πRH

= π(rl + r² + 2RH)

= 3.14 × (10 × 26 + 10² + 2 × 5 × 36)

= 3.14 × (260 + 100 + 360)

= 3.14 × 720

= 2260.8 cm².

Hence, total surface area of troy = 2260.8 cm².

Given,

Breadth of tunnel (b) = 6 m

Height of tunnel (h) = 8 m

Length of tunnel (l) = 35 m

Let radius of semi-circle be r meters.

From figure,

⇒ 2r = 6

⇒ r = 3 m.

Circumference of semi-circle = πr = $\frac{22}{7} \times 3 = \frac{66}{7}$ m.

Internal surface area of tunnel = Circumference of semi-circle × Length + Area of side interior rectangular walls

= πrl + hl + hl

= πrl + 2hl

= $\frac{66}{7} \times 35 + 2 \times 8 \times 35$

= 330 + 560

= 890 m².

Given,

Rate of plastering = ₹ 2.25 per m².

Total cost = 890 × ₹ 2.25 = ₹ 2002.50.

Hence, cost of plastering internal surface area of tunnel = ₹ 2002.50

To prove:

Equation : $\Big(\frac{\text{1 + cos A}}{\text{sin A}}\Big)^2 = \frac{\text{1 + cos A}}{\text{1 - cos A}}$.

Solving L.H.S. of the above equation :

$$\begin{aligned}\Rightarrow \Big(\frac{\text{1 + cos A}}{\text{sin A}}\Big)^2 \\[1em] \Rightarrow \frac{(\text{1 + cos A})^2}{\text{sin}^2 A} \\[1em] \Rightarrow \frac{(\text{1 + cos A})^2}{(\text{1 - cos}^2 A)} \\[1em] \Rightarrow \frac{(\text{1 + cos A})^2}{\text{(1 - cos A)(1 + cos A)}} \\[1em] \Rightarrow \frac{\text{1 + cos A}}{\text{1 - cos A}}.\end{aligned}$$

∴ L.H.S. = R.H.S.

Hence, proved that $\Big(\frac{\text{1 + cos A}}{\text{sin A}}\Big)^2 = \frac{\text{1 + cos A}}{\text{1 - cos A}}$.

To prove:

Equation : $\frac{\text{1 + sec A}}{\text{sec A}} = \frac{\text{sin}^2 A}{\text{1 - cos A}}$.

Solving L.H.S. of the above equation :

$$\begin{aligned}\Rightarrow \frac{\text{1 + sec A}}{\text{sec A}} \\[1em] \Rightarrow \frac{1 + \frac{1}{\text{cos A}}}{\frac{1}{\text{cos A}}} \\[1em] \Rightarrow \frac{\frac{\text{1 + cos A}}{\text{cos A}}}{\frac{1}{\text{cos A}}} \\[1em] \Rightarrow \frac{\text{1 + cos A}}{\text{cos A}} \times \text{cos A} \\[1em] \Rightarrow \text{1 + cos A}.\end{aligned}$$

Solving R.H.S. of the above equation :

$$\begin{aligned}\Rightarrow \frac{\text{sin}^2 A}{\text{1 - cos A}} \\[1em] \Rightarrow \frac{\text{1 - cos}^2 A}{\text{1 - cos A}} \\[1em] \Rightarrow \frac{\text{(1 - cos A)(1 + cos A)}}{\text {1 - cos A}} \\[1em] \Rightarrow \text{1 + cos A}.\end{aligned}$$

Since, L.H.S. = R.H.S = 1 + cos A.

Hence, proved that $\frac{\text{1 + sec A}}{\text{sec A}} = \frac{\text{sin}^2 A}{\text{1 - cos A}}$.

To prove:

Equation : sin A(1 + tan A) + cos A(1 + cot A) = sec A + cosec A.

Solving L.H.S. of the above equation :

$$\begin{aligned}\Rightarrow \text{sin A(1 + tan A) + cos A(1 + cot A)} \\[1em] \Rightarrow \text{sin A + sin A tan A + cos A + cos A cot A} \\[1em] \Rightarrow \text{sin A + sin A} \times \frac{\text{sin A}}{\text{cos A}} + \text{cos A + cos A} \times \frac{\text{cos A}}{\text{sin A}} \\[1em] \Rightarrow \text{sin A + cos A} + \frac{\text{sin}^2 A}{\text{cos A}} + \frac{\text{cos}^2 A}{\text{sin A}} \\[1em] \Rightarrow \Big(\text{sin A} + \frac{\text{cos}^2 A}{\text{sin A}}\Big) + \Big(\text{cos A} + \frac{\text{sin}^2 A}{\text{cos A}}\Big) \\[1em] \Rightarrow \Big(\frac{\text{sin}^2 A + \text{cos}^2 A}{\text{sin A}}\Big) + \Big(\frac{\text{cos}^2 A + \text{sin}^2 A}{\text{cos A}}\Big)\end{aligned}$$

By formula,

sin² A + cos² A = 1.

$$\begin{aligned}\Rightarrow \frac{1}{\text{sin A}} + \frac{1}{\text{cos A}} \\[1em] \Rightarrow \text{cosec A + sec A}.\end{aligned}$$

∴ L.H.S. = R.H.S.

Hence, proved that sin A(1 + tan A) + cos A(1 + cot A) = sec A + cosec A.

To prove:

Equation : sin² A tan A + cos² A cot A + 2 sin A cos A = tan A + cot A.

Solving L.H.S. of the above equation :

$$\begin{aligned}\Rightarrow \text{sin}^2 A \text{ tan A + cos}^2 A \text{ cot A + 2 sin A cos A} \\[1em] \Rightarrow \text{sin}^2 A \times \frac{\text{sin A}}{\text{cos A}} + \text{cos}^2 A \times \frac{\text{cos A}}{\text{sin A}} + \text{2 sin A cos A} \\[1em] \Rightarrow \frac{\text{sin}^3 A}{\text{cos A}} + \frac{\text{cos}^3 A}{\text{sin A}} + \text{2 sin A cos A} \\[1em] \Rightarrow \frac{\text{sin}^4 A + \text{cos}^4 A + \text{2 sin}^2 A \text{cos}^2 A}{\text{sin A cos A}} \\[1em] \Rightarrow \frac{(\text{sin}^2 A + \text{cos}^2 A)^2}{\text{sin A cos A}}\end{aligned}$$

By formula,

sin² A + cos² A = 1

$\Rightarrow \frac{1}{\text{sin A cos A}}$.

Solving R.H.S. of the above equation :

$$\begin{aligned}\Rightarrow \text{tan A + cot A} \\[1em] \Rightarrow \frac{\text{sin A}}{\text{cos A}} + \frac{\text{cos A}}{\text{sin A}} \\[1em] \Rightarrow \frac{\text{sin}^2 A + \text{cos}^2 A}{\text{cos A sin A}} \\[1em] \Rightarrow \frac{1}{\text{sin A cos A}}.\end{aligned}$$

Since, L.H.S. = R.H.S. = $\frac{1}{\text{sin A cos A}}$.

Hence, proved that sin² A tan A + cos² A cot A + 2 sin A cos A = tan A + cot A.

Solving R.H.S. of the equation :

$$\begin{aligned}\Rightarrow \frac{1}{\text{cosec A - cot A}} \\[1em] \Rightarrow \frac{1}{\frac{1}{\text{sin A}} - \frac{\text{cos A}}{\text{sin A}}} \\[1em] \Rightarrow \frac{1}{\frac{\text{1 - cos A}}{\text{sin A}}} \\[1em] \Rightarrow \frac{\text{sin A}}{\text{1 - cos A}}\end{aligned}$$

Multiplying numerator and denominator by (1 + cos A) we get :

$$\begin{aligned}\Rightarrow \frac{\text{sin A}}{\text{1 - cos A}} \times \frac{\text{1 + cos A}}{\text{1 + cos A}} \\[1em] \Rightarrow \frac{\text{sin A + sin A. cos A}}{\text{1 - cos}^2 A}\end{aligned}$$

By formula,

1 - cos² A = sin² A

$$\begin{aligned}\Rightarrow \frac{\text{sin A + sin A. cos A}}{\text{sin}^2 A} \\[1em] \Rightarrow \frac{\text{sin A}}{\text{sin}^2 A} + \frac{\text{sin A. cos A}}{\text{sin}^2 A} \\[1em] \Rightarrow \frac{1}{\text{sin A}} + \frac{\text{cos A}}{\text{sin A}} \\[1em] \Rightarrow \text{cosec A + cot A}.\end{aligned}$$

Since, L.H.S. = R.H.S.

Hence, proved that cosec A + cot A = $\frac{1}{\text{cosec A - cot A}}$.

Solving L.H.S. of the equation :

$\Rightarrow \frac{\text{cot A + cosec A - 1}}{\text{cot A - cosec A + 1}}$

By formula,

cosec² A - cot² A = 1

$$\begin{aligned}\Rightarrow \frac{\text{cot A + cosec A - (cosec}^2 A - \text{cot}^2 A)}{\text{cot A - cosec A + 1}} \\[1em] \Rightarrow \frac{\text{(cot A + cosec A) - (cosec} A - \text{cot} A)\text{(cosec A + cot A)}}{\text{cot A - cosec A + 1}} \\[1em] \Rightarrow \frac{\text{(cot A + cosec A)[1 - (cosec A - cot A)]}}{\text{cot A - cosec A + 1}} \\[1em] \Rightarrow \frac{\text{(cot A + cosec A)(cot A - cosec A + 1)}}{\text{cot A - cosec A + 1}} \\[1em] \Rightarrow \text{cot A + cosec A} \\[1em] \Rightarrow \frac{\text{cos A}}{\text{sin A}} + \frac{1}{\text{sin A}} \\[1em] \Rightarrow \frac{\text{cos A + 1}}{\text{sin A}}.\end{aligned}$$

Since, L.H.S. = R.H.S.

Hence, proved that $\frac{\text{cot A + cosec A - 1}}{\text{cot A - cosec A + 1}} = \frac{\text{1 + cos A}}{\text{sin A}}$.

Given,

⇒ tan A = 1

⇒ tan A = tan 45°

⇒ A = 45°.

⇒ tan B = $\sqrt{3}$

⇒ tan B = tan 60°

⇒ B = 60°.

(i) Solving,

⇒ cos A cos B - sin A sin B

⇒ cos 45° cos 60° - sin 45° sin 60°

$$\begin{aligned}\Rightarrow \frac{1}{\sqrt{2}} \times \frac{1}{2} - \frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2} \\[1em] \Rightarrow \frac{1}{2\sqrt{2}} - \frac{\sqrt{3}}{2\sqrt{2}} \\[1em] \Rightarrow \frac{1 - \sqrt{3}}{2\sqrt{2}} \\[1em] \Rightarrow \frac{1 - \sqrt{3}}{2\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} \\[1em] \Rightarrow \frac{\sqrt{2} - \sqrt{3}\sqrt{2}}{4} \\[1em] \Rightarrow \frac{\sqrt{2} - \sqrt{6}}{4}.\end{aligned}$$

Hence, cos A cos B - sin A sin B = $\frac{\sqrt{2} - \sqrt{6}}{4}.$

(ii) Given,

⇒ sin A cos B + cos A sin B

⇒ sin 45° cos 60° + cos 45° sin 60°

$$\begin{aligned}\Rightarrow \frac{1}{\sqrt{2}} \times \frac{1}{2} + \frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2} \\[1em] \Rightarrow \frac{1}{2\sqrt{2}} + \frac{\sqrt{3}}{2\sqrt{2}} \\[1em] \Rightarrow \frac{1 + \sqrt{3}}{2\sqrt{2}} \\[1em] \Rightarrow \frac{1 + \sqrt{3}}{2\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} \\[1em] \Rightarrow \frac{\sqrt{2}(1 + \sqrt{3})}{4} \\[1em] \Rightarrow \frac{\sqrt{2} + \sqrt{6}}{4}.\end{aligned}$$

Hence, sin A cos B + cos A sin B = $\frac{\sqrt{2} + \sqrt{6}}{4}.$

Let AB be the lighthouse and C and D be the position of ships.

From figure,

∠ADC = ∠EAD = 30° (Alternate angles are equal)

∠ACB = ∠EAC = 45° (Alternate angles are equal)

In right angle triangle ABD,

⇒ tan 30° = $\frac{\text{Perpendicular}}{\text{Base}}$

⇒ $\frac{1}{\sqrt{3}} = \frac{AB}{BD}$

⇒ BD = $\sqrt{3}AB = 100\sqrt{3}$ m.

In right angle triangle ABC,

⇒ tan 45° = $\frac{\text{Perpendicular}}{\text{Base}}$

⇒ $1 = \frac{AB}{BC}$

⇒ BC = AB = 100 m.

From figure,

CD = BD - BC = $100\sqrt{3} - 100 = 100(\sqrt{3} - 1)$

= 100 × (1.732 - 1) m.

= 100 × 0.732 = 73.2 m.

Hence, distance between two ships = 73.2 m.

Given,

⇒ (2 cos 2A - 1)(tan 3A - 1) = 0

⇒ 2 cos 2A - 1 = 0 or tan 3A - 1 = 0

⇒ 2 cos 2A = 1 or tan 3A = 1

⇒ cos 2A = $\frac{1}{2}$ or tan 3A = tan 45°

⇒ cos 2A = cos 60° or tan 3A = tan 45°

⇒ 2A = 60° or 3A = 45°

⇒ A = $\frac{60°}{2}$ or A = $\frac{45°}{3}$

⇒ A = 30° or 15°.

Hence, A = 30° or 15°.

Given,

cos A = $\frac{9}{41}$;

Squaring both sides, we get :

$$\begin{aligned}\Rightarrow \text{cos}^2 A = \Big(\frac{9}{41}\Big)^2 \\[1em] \Rightarrow \text{cos}^2 A = \frac{81}{1681} \\[1em] \Rightarrow \text{1 - sin}^2 A = \frac{81}{1681} \\[1em] \Rightarrow \text{sin}^2 A = 1 - \frac{81}{1681} \\[1em] \Rightarrow \text{sin}^2 A = \frac{1681 - 81}{1681} \\[1em] \Rightarrow \text{sin}^2 A = \frac{1600}{1681} \\[1em] \Rightarrow \text{sin A} = \sqrt{\frac{1600}{1681}} \\[1em] \Rightarrow \text{sin A} = \frac{40}{41}.\end{aligned}$$

By formula,

tan A = $\frac{\text{sin A}}{\text{cos A}} = \frac{\frac{40}{41}}{\frac{9}{41}} = \frac{40}{9}$.

Substituting value of sin A and tan A in $\frac{1}{\text{sin}^2 A} - \frac{1}{\text{tan}^2 A}$, we get :

$$\begin{aligned}\Rightarrow \frac{1}{\text{sin}^2 A} - \frac{1}{\text{tan}^2 A} \\[1em] \Rightarrow \frac{1}{\Big(\frac{40}{41}\Big)^2} - \frac{1}{\Big(\frac{40}{9}\Big)^2} \\[1em] \Rightarrow \Big(\frac{41}{40}\Big)^2 - \Big(\frac{9}{40}\Big)^2 \\[1em] \Rightarrow \frac{1681}{1600} - \frac{81}{1600} \\[1em] \Rightarrow \frac{1600}{1600} \\[1em] \Rightarrow 1.\end{aligned}$$

Hence, $\frac{1}{\text{sin}^2 A} - \frac{1}{\text{tan}^2 A}$ = 1.

Let AB be the lighthouse of height h meters and C and D be the position of the ship.

From figure,

∠ADC = ∠EAD = 30° (Alternate angles are equal)

∠ACB = ∠EAC = 45° (Alternate angles are equal)

From figure,

In right angle triangle ADB

⇒ tan 30° = $\frac{\text{Perpendicular}}{\text{Base}}$

⇒ $\frac{1}{\sqrt{3}} = \frac{AB}{BD}$

⇒ BD = $\sqrt{3}\text{AB} = \sqrt{3}h$.

In right angle triangle ABC

⇒ tan 45° = $\frac{\text{Perpendicular}}{\text{Base}}$

⇒ 1 = $\frac{AB}{BC}$

⇒ BC = AB = h.

From figure,

CD = BD - BC = $\sqrt{3}h - h = h(\sqrt{3} - 1)$ meters.

Given,

Angle of depression of the ship changes from 30° to 45° in 10 minutes.

∴ Ship comes from position D to C in 10 minutes.

∴ Ship covers a distance of CD = $h(\sqrt{3} - 1)$ in 10 minutes.

Speed of ship = $\frac{\text{Distance}}{\text{Time}} = \frac{h(\sqrt{3} - 1)}{10}$ meter/minute.

Time taken by ship to cover (BC = h meters) is

$$\begin{aligned}\text{Time} = \frac{\text{Distance}}{\text{Speed}} \\[1em] = \frac{h}{\frac{h(\sqrt{3} - 1)}{10}} \\[1em] = \frac{10}{\sqrt{3} - 1} \\[1em] = \frac{10}{1.732 - 1} \\[1em] = \frac{10}{0.732} \\[1em] = 13.66 \text{ minutes}.\end{aligned}$$

Hence, ship will reach the lighthouse in 13.66 minutes.

Let AB be the tower of height h meters, BC = a meters and BD = b meters.

According to question,

⇒ α + β = 90°

⇒ β = 90° - α

From figure,

In △ABD,

$$\begin{aligned}\Rightarrow \text{tan α} = \frac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow \text{tan α} = \frac{AB}{BD} \\[1em] \Rightarrow \text{tan α} = \frac{h}{b} ...........(1)\end{aligned}$$

In △ABC,

$$\begin{aligned}\Rightarrow \text{tan β} = \frac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow \text{tan β} = \frac{AB}{BC} \\[1em] \Rightarrow \text{tan β} = \frac{h}{a} \\[1em] \Rightarrow \text{tan (90° - α)} = \frac{h}{a} \\[1em] \Rightarrow \text{cot α} = \frac{h}{a} ..........(2)\end{aligned}$$

Multiplying (1) by (2) we get,

⇒ tan α cot α = $\frac{h}{a} \times \frac{h}{b}$

⇒ $\frac{\text{sin α}}{\text{cos α}} \times \frac{\text{cos α}}{\text{sin α}} = \frac{h^2}{ab}$

⇒ 1 = $\frac{h^2}{ab}$

⇒ h² = ab

⇒ h = $\sqrt{ab}$ meters.

Hence, proved that h = $\sqrt{ab}$ meters.

(i) Let CD be the tower and AB be the pole,

In △ABD,

$$\begin{aligned}\text{tan 30°} = \frac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow \frac{1}{\sqrt{3}} = \frac{AB}{BD} \\[1em] \Rightarrow BD = \sqrt{3}AB \\[1em] \Rightarrow BD = 1.732 \times 20 \\[1em] \Rightarrow BD = 34.64 \text{ m}.\end{aligned}$$

In △CBD,

$$\begin{aligned}\text{tan 60°} = \frac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow \sqrt{3} = \frac{CD}{BD} \\[1em] \Rightarrow CD = \sqrt{3}BD \\[1em] \Rightarrow CD = 1.732 \times 34.64 \\[1em] \Rightarrow CD = 60 \text{ m}.\end{aligned}$$

Hence, height of tower = 60 meters.

(ii) From part (i),

BD = 34.64 m

Hence, horizontal distance between pole and tower = 34.64 meters.

Let DE be the surface of lake and A be the point 36 m above surface of lake and F is the image of bird (B).

Let BC be h meters.

BD = DF (As distance of bird from lake = distance of image of bird in lake from lake)

In △ABC,

$$\begin{aligned}\text{tan 30°} = \frac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow \frac{1}{\sqrt{3}} = \frac{BC}{AC} \\[1em] \Rightarrow AC = \sqrt{3}BC \\[1em] \Rightarrow AC = \sqrt{3}h.\end{aligned}$$

From figure,

⇒ CF = CD + DF

⇒ CF = CD + BD

⇒ CF = CD + BC + CD

⇒ CF = 36 + h + 36

⇒ CF = h + 72.

In △ACF,

$$\begin{aligned}\text{tan 60°} = \frac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow \sqrt{3} = \frac{CF}{AC} \\[1em] \Rightarrow \sqrt{3} = \frac{h + 72}{\sqrt{3}h} \\[1em] \Rightarrow 3h = h + 72 \\[1em] \Rightarrow 2h = 72 \\[1em] \Rightarrow h = 36 \text{ m}.\end{aligned}$$

Height of bird from lake = BD = BC + CD = h + 36 = 36 + 36 = 72 m.

Hence, height of bird above surface of lake = 72 meters.

The above distribution is discontinuous converting into continuous distribution, we get :

Adjustment factor = (Lower limit of one class - Upper limit of previous class) / 2

= $\frac{21 - 20}{2} = \frac{1}{2}$

= 0.5

Subtract the adjustment factor (0.5) from all the lower limits and add the adjustment factor (0.5) to all the upper limits.

Marks	Frequency (f)	Cumulative frequency	Class mark (x)	fx
0.5 - 10.5	7	7	5.5	38.5
10.5 - 20.5	9	16 (9 + 7)	15.5	139.5
20.5 - 30.5	15	31 (15 + 16)	25.5	382.5
30.5 - 40.5	8	39 (31 + 8)	35.5	284
40.5 - 50.5	6	45 (39 + 6)	45.5	273
50.5 - 60.5	5	50 (45 + 5)	55.5	277.5
Total	Σf = 50			Σfx = 1395

Mean = $\frac{Σfx}{Σf} = \frac{1395}{50}$ = 27.9

Given, total terms = 50.

Since, terms are even.

By formula,

Median = $\frac{50}{2}$ = 25th term.

From table,

15.5 th to 25.5 th term marks lies in the range 20.5 - 30.5.

∴ Median class = 21 - 30.

From table class 21 - 30 has the highest frequency.

∴ Modal class = 21 - 30.

Hence, mean = 27.9, median class = 21 - 30 and modal class = 21 - 30.

Cumulative frequency distribution :

Marks (x)	Frequency (f)	Cumulative frequency	fx
3	3	3	9
4	5	8 (5 + 3)	20
5	8	16 (8 + 8)	40
6	7	23 (7 + 16)	42
7	9	32 (9 + 23)	63
8	4	36 (4 + 32)	32
9	3	39 (3 + 36)	27
10	1	40 (1 + 39)	10
Total	Σf = 40		Σfx = 243

By formula,

Mean = $\frac{Σfx}{Σf} = \frac{243}{40}$ = 6.075

Given,

Total terms = 40, which is even.

By formula,

Median = $\frac{n}{2} = \frac{40}{2}$ = 20th term.

From table,

Marks of 17th to 23rd term = 6.

∴ Median = 6.

Highest frequency is of 7 marks.

∴ Mode = 7.

Hence, mean = 6.075, median = 6 and mode = 7.

Cumulative frequency distribution table :

Weight (kg)	No. of boys (frequency)	Cumulative frequency
10 - 15	11	11
15 - 20	25	36 (11 + 25)
20 - 25	12	48 (36 + 12)
25 - 30	5	53 (48 + 5)
30 - 35	2	55 (53 + 2)

Here, n = 55, which is odd.

By formula,

Median = $\frac{n + 1}{2}\text{ th term} = \frac{55 + 1}{2} = \frac{56}{2}$ = 28th term.

Steps of construction of ogive :

1. Take 2 cm = 5 units on x-axis.

2. Take 1 cm = 10 units on y-axis.

3. Plot the point (10, 0), as ogive always starts on x-axis representing the lower limit of the first class.

4. Plot the points (15, 11), (20, 36), (25, 48), (30, 53) and (35, 55).

5. Join the points by a free hand curve.

6. Draw a line parallel to x-axis from point A (frequency) = 28, touching the graph at point B. From point B draw a line parallel to y-axis touching x-axis at point C.

From graph, C = 18.4

Hence, median = 18.4

Cumulative frequency distribution table :

Class	Frequency	Cumulative frequency
0 - 10	4	4
10 - 20	9	13 (4 + 9)
20 - 30	15	28 (13 + 15)
30 - 40	14	42 (28 + 14)
40 - 50	8	50 (42 + 8)

Here, n = 50, which is even.

By formula,

Median = $\frac{n}{2}\text{ th term} = \frac{50}{2}$ = 25th term.

Steps of construction of ogive :

1. Take 2 cm = 10 units on x-axis.

2. Take 1 cm = 5 units on y-axis.

3. Plot the point (0, 0), as ogive always starts on x-axis representing the lower limit of the first class.

4. Plot the points (10, 4), (20, 13), (30, 28), (40, 42) and (50, 50).

5. Join the points by a free hand curve.

6. Draw a line parallel to x-axis from point A (frequency) = 25, touching the graph at point B. From point B draw a line parallel to y-axis touching x-axis at point C.

From graph, C = 28.

Hence, median = 28.

(i) Cumulative frequency distribution table :

Height (in cm)	No. of students	Cumulative frequency
140 - 145	12	12
145 - 150	20	32 (12 + 20)
150 - 155	30	62 (32 + 30)
155 - 160	38	100 (62 + 38)
160 - 165	24	124 (100 + 24)
165 - 170	16	140 (124 + 16)
170 - 175	12	152 (140 + 12)
175 - 180	8	160 (152 + 8)

Here, n = 160, which is even.

Median = $\frac{n}{2}$ th term = $\frac{160}{2}$ = 80th term.

Steps of construction :

1. Take 1 cm = 5 cm on x-axis.

2. Take 1 cm = 20 students on y-axis.

3. Since, x axis starts at 140 hence, a kink is drawn at the starting of x-axis. Plot the point (140, 0) as ogive starts on x-axis representing lower limit of first class.

4. Plot the points (145, 12), (150, 32), (155, 62), (160, 100), (165, 124), (170, 140), (175, 152) and (180, 160).

5. Join the points by a free-hand curve.

6. Draw a line parallel to x-axis from point J (no. of students) = 80, touching the graph at point K. From point K draw a line parallel to y-axis touching x-axis at point L.

From graph, L = 157.5

Hence, median = 157.5 cm

(ii) Here, n = 160, which is even.

By formula,

Lower quartile = $\frac{n}{4} = \frac{160}{4}$ = 40th term.

Draw a line parallel to x-axis from point P (no. of students) = 40, touching the graph at point Q. From point Q draw a line parallel to y-axis touching x-axis at point R.

From graph, R = 152

Upper quartile = $\frac{3n}{4} = \frac{3 \times 160}{4}$ = 120th term.

Draw a line parallel to x-axis from point M (no. of students) = 120, touching the graph at point N. From point N draw a line parallel to y-axis touching x-axis at point O.

From graph, O = 164

Inter quartile range = Upper quartile - Lower quartile

= 164 - 152 = 12.

Hence, inter quartile range = 12.

(iii) Draw a line parallel to y-axis from point S (height) = 172 cm, touching the graph at point T. From point T draw a line parallel to x-axis touching y-axis at point U.

From graph, U = 144.

∴ 144 students have height less than or equal to 172 cm.

No. of students whose height is more than 172 cm = 160 - 144 = 16.

Hence, no. of students whose height is more than 172 cm = 16.

(i) The above distribution is discontinuous, converting into continuous distribution, we get :

Adjustment factor = (Lower limit of one class - Upper limit of previous class) / 2

= $\frac{20 - 19}{2} = \frac{1}{2}$

= 0.5

Subtract the adjustment factor (0.5) from all the lower limits and add the adjustment factor (0.5) to all the upper limits.

Cumulative frequency distribution table :

Marks	No. of students	Cumulative frequency distribution
9.5 - 19.5	14	14
19.5 - 29.5	16	30 (14 + 16)
29.5 - 39.5	22	52 (30 + 22)
39.5 - 49.5	26	78 (52 + 26)
49.5 - 59.5	18	96 (78 + 18)
59.5 - 69.5	11	107 (96 + 11)
69.5 - 79.5	6	113 (107 + 6)
79.5 - 89.5	4	117 (113 + 4)
89.5 - 99.5	3	120 (117 + 3)

Here, n = 120, which is even.

Median = $\frac{n}{2}$th term

= $\frac{120}{2}$ = 60th term.

Steps of construction :

1. Since, the scale on x-axis starts at 9.5, a break (kink) is shown near the origin on x-axis to indicate that the graph is drawn to scale beginning at 9.5

2. Take 1 cm along x-axis = 10 units.

3. Take 1 cm along y-axis = 20 units.

4. Plot the point (9.5, 0) as ogive starts on x-axis representing lower limit of first class.

5. Plot the points (19.5, 14), (29.5, 30), (39.5, 52), (49.5, 78), (59.5, 96), (69.5, 107), (79.5, 113), (89.5, 117) and (99.5, 120).

6. Join the points by a free-hand curve.

7. Draw a line parallel to x-axis from point O (no. of students) = 60, touching the graph at point P. From point P draw a line parallel to y-axis touching x-axis at point Q.

From graph, Q = 42.5 or 43

Hence, median = 42.5 or 43.

(ii) Total marks = 100.

75% marks = $\frac{75}{100} \times 100$ = 75.

Draw a line parallel to y-axis from point R (marks) = 75, touching the graph at point S. From point S draw a line parallel to x-axis touching y-axis at point T.

From graph, T = 111.

∴ 111 students score less than or equal to 75%.

Students scoring above than 75% = 120 - 111 = 9.

Hence, 9 students score more than 75%.

(i) Cumulative frequency distribution table :

Income (in thousand ₹)	No. of students(frequency)	Cumulative frequency
0 - 8	8	8
8 - 16	35	43 (8 + 35)
16 - 24	35	78 (43 + 35)
24 - 32	14	92 (78 + 14)
32 - 40	8	100 (92 + 8)

Here, n = 100, which is even.

Median = $\frac{n}{2} = \frac{100}{2}$ = 50th term.

Steps of construction of ogive :

1. Take 2 cm = 8 thousand on x-axis.

2. Take 1 cm = 10 students on y-axis.

3. Plot the point (0, 0) as ogive starts from x-axis representing the lower limit of first class.

4. Plot the points (8, 8), (16, 43), (24, 78), (32, 92) and (40, 100).

5. Join the points by a free hand curve.

6. Draw a line parallel to x-axis from point G (no. of students) = 50, touching the graph at point H. From point H draw a line parallel to y-axis touching x-axis at point I.

From graph, I = 17.6 (thousands)

Hence, median = 17.6 thousands.

(ii) 15% of 100 students = $\frac{15}{100} \times 100$ = 15.

Draw a line parallel to x-axis from point J (no. of students) = 15, touching the graph at point K. From point K draw a line parallel to y-axis touching x-axis at point L.

From graph,

L = 9.6 thousands.

Hence, the annual income of parents, below which the freeships will be awarded is 9.6 thousands.

(iii)

Income (in thousand ₹)	Class mean (x)	No. of students (f)	fx
0 - 8	4	8	32
8 - 16	12	35	420
16 - 24	20	35	700
24 - 32	28	14	392
32 - 40	36	8	288
Total		Σf = 100	Σfx = 1832

Mean = $\frac{Σfx}{Σf} = \frac{1832}{100}$ = 18.32 (thousands)

Hence, mean = 18.32 (thousands).

Cumulative frequency distribution table :

Income in ₹	No. of employees	Cumulative frequency
120 - 140	30	30
140 - 160	72	102 (30 + 72)
160 - 180	90	192 (102 + 90)
180 - 200	80	272 (192 + 80)
200 - 220	70	342 (272 + 70)
220 - 240	28	370 (342 + 28)

(i) Here, n = 370, which is even,

Median = $\frac{n}{2} = \frac{370}{2}$ = 185.

Steps of construction :

1. Take 2 cm = ₹ 20 on x-axis.

2. Take 1 cm = 50 employees on y-axis.

3. A kink is shown near x-axis as it starts from 120. Plot the point (120, 0) as ogive starts on x-axis representing lower limit of first class.

4. Plot the points (140, 30), (160, 102), (180, 192), (200, 272), (220, 342) and (240, 370).

5. Join the points by a free-hand curve.

6. Draw a line parallel to x-axis from point H (no. of employees) = 185, touching the graph at point I. From point I draw a line parallel to y-axis touching x-axis at point J.

From graph, J = 179

Hence, median wage = ₹ 179.

(ii) Steps :

1. Draw a line parallel to y-axis from point K (income) = ₹ 190, touching the graph at point L. From point L draw a line parallel to x-axis touching y-axis at point M.

From graph, M = 232

It means that,

232 employees have income less than or equal to ₹ 190.

∴ No. of employees having income more than ₹ 190 = 370 - 232 = 138.

Hence, no. of employees having income more than ₹ 190 = 138.

The above table can be written as :

Marks	No. of candidates
0 - 10	0
10 - 20	25
20 - 30	42
30 - 40	65
40 - 50	95
50 - 60	120
60 - 70	128
70 - 80	135
80 - 90	148
90 - 100	150

Steps of construction :

1. Take 1 cm = 10 marks on x-axis.

2. Take 1 cm = 20 students on y-axis.

3. Plot the point (0, 0) as ogive starts on x-axis representing lower limit of first class.

4. Plot the points (10, 0), (20, 25), (30, 42), (40, 65), (50, 95), (60, 120), (70, 128), (80, 135), (90, 148) and (100, 150).

5. Join the points by a free-hand curve.

6. Draw a line parallel to y-axis from point L (marks) = 45, touching the graph at point M. From point M draw a line parallel to x-axis touching y-axis at point N.

From graph, N = 80

Hence, no. of students having marks less than 45 = 80.

(ii) Steps :

1. Draw a line parallel to y-axis from point O (marks) = 75, touching the graph at point P. From point P draw a line parallel to x-axis touching y-axis at point Q.

From graph, Q = 131

It means that,

131 students have marks less than or equal to 75.

∴ No. of students having marks more than 75 = 150 - 131 = 19.

Hence, no. of students having marks more than 75 = 19.

(i) Cumulative frequency distribution table :

Marks	No. of students	Cumulative frequency
0 - 10	5	5
10 - 20	11	16
20 - 30	10	26
30 - 40	20	46
40 - 50	28	74
50 - 60	37	111
60 - 70	40	151
70 - 80	29	180
80 - 90	14	194
90 - 100	6	200

Here, n = 200, which is even.

By formula,

Median = $\frac{n}{2}$ th term

= $\frac{200}{2}$ = 100th term.

Steps of construction :

1. Take 1 cm along x-axis = 10 marks.

2. Take 1 cm along y-axis = 20 students.

3. Plot the point (0, 0) as ogive starts from x-axis representing lower limit of first class.

4. Plot the points (10, 5), (20, 16), (30, 26), (40, 46), (50, 74), (60, 111), (70, 151), (80, 180), (90, 194) and (100, 200).

5. Join the points by a free hand curve.

6. Draw a line parallel to x-axis from point L (no. of students) = 100, touching the graph at point M. From point M draw a line parallel to y-axis touching x-axis at point N.

From graph, N = 57

Hence, median = 57.

(ii) Minimum marks required to pass = 40.

Draw a line parallel to y-axis from point O (marks) = 40, touching the graph at point Q. From point Q draw a line parallel to x-axis touching y-axis at point P.

From graph,

P = 46.

∴ 46 students scored less than 40 marks.

Hence, 46 students failed the examination.

(iii) 85 or more marks is considered grade one.

Draw a line parallel to y-axis from point Q (marks) = 85, touching the graph at point R. From point R draw a line parallel to x-axis touching y-axis at point S.

From graph,

S = 188.

∴ 188 students scored less than 85 marks.

∴ 12 (200 - 188) students scored more than 85 marks.

Hence, 12 students secured grade one.

Marks (x)	Number of students (f)	fx
5	6	30
6	a	6a
7	16	112
8	13	104
9	b	9b
Total	Σf = a + b + 35	Σfx = 246 + 6a + 9b

Total 40 students.

∴ a + b + 35 = 40

⇒ a + b = 40 - 35

⇒ a + b = 5

⇒ a = 5 - b ........(1)

Given, mean = 7.2

$$\begin{aligned}\therefore 7.2 = \frac{Σfx}{Σf} \\[1em] \Rightarrow 7.2 = \frac{246 + 6a + 9b}{40}\end{aligned}$$

Substituting value of a from (1) in above equation :

$$\begin{aligned}\Rightarrow 7.2 = \frac{246 + 6(5 - b) + 9b}{40} \\[1em] \Rightarrow 7.2 \times 40 = 246 + 30 - 6b + 9b \\[1em] \Rightarrow 288 = 276 + 3b \\[1em] \Rightarrow 3b = 288 - 276 \\[1em] \Rightarrow 3b = 12 \\[1em] \Rightarrow b = \frac{12}{3} \\[1em] \Rightarrow b = 4.\end{aligned}$$

a = 5 - b = 5 - 4 = 1.

Hence, a = 1 and b = 4.

(i) Cumulative frequency distribution table :

Marks	No. of students	Cumulative frequency
0 - 10	3	3
10 - 20	7	10
20 - 30	12	22
30 - 40	17	39
40 - 50	23	62
50 - 60	14	76
60 - 70	9	85
70 - 80	6	91
80 - 90	5	96
90 - 100	4	100

Here, n = 100, which is even.

Median = $\frac{n}{2}$ th term

= $\frac{100}{2}$ = 50th term.

Steps of construction of ogive :

1. Take 1 cm = 10 marks on x-axis.

2. Take 1 cm = 20 students on y-axis.

3. Plot the point (0, 0) as ogive starts from x-axis representing lower limit of first class.

4. Plot the points (10, 3), (20, 10), (30, 22), (40, 39), (50, 62), (60, 76), (70, 85), (80, 91), (90, 96) and (100, 100).

5. Join the points by a free hand curve.

6. Draw a line parallel to x-axis from point L (no. of students) = 50, touching the graph at point M. From point M draw a line parallel to y-axis touching x-axis at point N.

From graph, N = 45

Hence, median = 45.

(ii) Here, n = 100, which is even.

By formula,

Lower quartile = $\frac{n}{4} = \frac{100}{4}$ = 25th term.

Draw a line parallel to x-axis from point O (no. of students) = 25, touching the graph at point P. From point P draw a line parallel to y-axis touching x-axis at point Q.

From graph, Q = 32

Hence, lower quartile = 32.

(iii) Total marks = 100.

85% of 100 marks = $\frac{85}{100} \times 100$ = 85.

Draw a line parallel to y-axis from point R (marks) = 85, touching the graph at point S. From point S draw a line parallel to x-axis touching y-axis at point T.

From graph, T = 94.

It means that 94 students scored either less or equal to 85% marks.

No. of students left = 100 - 94 = 6.

Hence, no. of students scoring more than 85% marks = 6.

(iv) Total marks = 100.

35% of 100 marks = $\frac{35}{100} \times 100$ = 35.

Draw a line parallel to y-axis from point U (marks) = 35, touching the graph at point V. From point V draw a line parallel to x-axis touching y-axis at point W.

From graph, W = 28.

Hence, 28 students fail if pass percentage was 35%.

No. of possible outcomes = 3 (1 red + 1 blue + 1 yellow)

(i) No. of yellow balls = 1.

∴ No. of favourable outcomes = 1

P(drawing a yellow ball) = $\frac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \frac{1}{3}$.

Hence, probability of drawing a yellow ball = $\frac{1}{3}$.

(ii) No. of red balls = 1.

∴ No. of favourable outcomes = 1

P(drawing a red ball) = $\frac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \frac{1}{3}$.

Hence, probability of drawing a red ball = $\frac{1}{3}$.

(iii) No. of blue balls = 1.

∴ No. of favourable outcomes = 1

P(drawing a blue ball) = $\frac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \frac{1}{3}$.

Hence, probability of drawing a blue ball = $\frac{1}{3}$.

No. of possible outcomes = 24 (6 red + 8 blue + 10 yellow)

(i) No. of yellow balls = 10.

∴ No. of favourable outcomes = 10

P(drawing a yellow ball) = $\frac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \frac{10}{24} = \frac{5}{12}$.

Hence, probability of drawing a yellow ball = $\frac{5}{12}$.

(ii) No. of red balls = 6.

∴ No. of favourable outcomes = 6

P(drawing a red ball) = $\frac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \frac{6}{24} = \frac{1}{4}$.

Hence, probability of drawing a red ball = $\frac{1}{4}$.

(iii) No. of blue balls = 8.

∴ No. of favourable outcomes = 8

P(drawing a blue ball) = $\frac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \frac{8}{24} = \frac{1}{3}$.

Hence, probability of drawing a blue ball = $\frac{1}{3}$.

(iv) No. of balls that are not yellow = 14 (6 red + 8 blue)

∴ No. of favourable outcomes = 14

P(not drawing a yellow ball) = $\frac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \frac{14}{24} = \frac{7}{12}$.

Hence, probability of not drawing a yellow ball = $\frac{7}{12}$.

(v) No. of balls that are not blue = 16 (6 red + 10 yellow)

∴ No. of favourable outcomes = 16

P(not drawing a blue ball) = $\frac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \frac{16}{24} = \frac{2}{3}$.

Hence, probability of not drawing a blue ball = $\frac{2}{3}$.

When two dice are thrown, simultaneously.

Possible outcomes = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}.

No. of possible outcomes = 6 × 6 = 36.

(i) Favourable outcomes for getting a sum of 13 = 0.

∴ No. of favourable outcomes = 0

P(getting a sum of 13) = $\frac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \frac{0}{36}$ = 0.

Hence, probability of getting a sum of 13 = 0.

(ii) Favourable outcomes for getting a sum less than 13 = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}.

∴ No. of favourable outcomes = 36

P(getting a sum less than 13) = $\frac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \frac{36}{36}$ = 1.

Hence, probability of getting a sum less than 13 = 1.

(iii) Favourable outcomes for getting a sum of 10 = {(4, 6), (5, 5), (6, 4)}.

∴ No. of favourable outcomes = 3

P(getting a sum of 10) = $\frac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \frac{3}{36} = \frac{1}{12}$.

Hence, probability of getting a sum of 10 = $\frac{1}{12}$.

(iv) Favourable outcomes for getting a sum less than 10 = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (5, 1), (5, 2), (5, 3), (5, 4), (6, 1), (6, 2), (6, 3)}.

∴ No. of favourable outcomes = 30

P(getting a sum less than 10) = $\frac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \frac{30}{36} = \frac{5}{6}$.

Hence, probability of getting a sum less than 10 = $\frac{5}{6}$.

There are five cards.

∴ No. of possible outcomes = 5.

(i) No. of queens = 1

∴ No. of favourable outcomes = 1

P(getting a queen) = $\frac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \frac{1}{5}$.

Hence, probability of drawing a queen = $\frac{1}{5}$.

(ii) Since, queen is drawn and put aside.

∴ No. of cards left = 5 - 1 = 4

∴ No. of possible outcomes = 4.

(a) No. of ace cards = 1

∴ No. of favourable outcomes = 1

P(getting an ace) = $\frac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \frac{1}{4}$.

Hence, probability of drawing an ace = $\frac{1}{4}$.

(b) No. of queen left = 0

∴ No. of favourable outcomes = 0

P(getting a queen) = $\frac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \frac{0}{4}$ = 0.

Hence, probability of getting a queen = 0.

No. of defective bulbs = 4.

∴ No. of favourable outcomes = 4

P(getting a defective ball) = $\frac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \frac{4}{20} = \frac{1}{5}$.

Hence, probability of getting a defective bulb = $\frac{1}{5}$.

(ii) Since, one bulb is drawn at random and is not defective and is not replaced.

∴ No. of possible outcomes = 19.

∴ No. of good bulbs present = 15 (16 - 1)

∴ No. of favourable outcomes = 15

P(not getting a defective ball) = $\frac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \frac{15}{19}$.

Hence, probability of not getting a defective bulb = $\frac{15}{19}$.